1.1.2.1.
Answer.
Appendix E Answers to Exercises
1 Limits
1.1 Drawing Tangents and a First Limit
1.1.2 Exercises
1.1.2.2.
Answer.
- True
- In general, this is false. For most functions \(f(x)\) this will be false, but there are some functions for which it is true.
1.1.2.3.
Answer.
At least once.
1.2 Another Limit and Computing Velocity
1.2.2 Exercises
1.2.2.1.
Answer.
Speed is nonnegative; velocity has a sign (positive or negative) that indicates direction.
1.2.2.2.
Answer.
Yes--an object that is not moving has speed 0.
1.2.2.3.
Answer.
0 kph
1.2.2.4.
Answer.
The speed at the one second mark is larger than the average speed.
1.2.2.5.
Answer.
The slope of a curve is given by \(\dfrac{\mbox{change in vertical component}}{\mbox{change in horizontal component}}\text{.}\) The change in the vertical component is exactly \(s(b)-s(a)\text{,}\) and the change in the horizontal component is exactly \(b-a\text{.}\)
1.2.2.6.
Answer.
\((0,2) \cup (6,7)\)
1.2.2.7.
Answer.
- 24 units per second.
- 6 units per second
1.2.2.8.
Answer.
- \(\frac{1}{4}\) units per second
- \(\frac{1}{2}\) units per second
- \(\frac{1}{6}\) units per second
Remark: the average velocity is not the average of the two instantaneous velocities.
1.3 The Limit of a Function
1.3.2 Exercises
1.3.2.1.
Answer.
- \(\displaystyle \displaystyle \lim_{x \rightarrow -2} f(x)=1\)
- \(\displaystyle \displaystyle \lim_{x \rightarrow 0}f(x)=0\)
- \(\displaystyle \displaystyle \lim_{x \rightarrow 2}f(x)=2\)
1.3.2.2.
Answer.
DNE
1.3.2.3.
Answer.
- \(\displaystyle \displaystyle \lim_{x \rightarrow -1^{-}} f(x)=2\)
- \(\displaystyle \displaystyle \lim_{x \rightarrow -1^{+}} f(x)=-2\)
- \(\displaystyle \lim_{x \rightarrow -1} f(x)= \) DNE
- \(\displaystyle \displaystyle \lim_{x \rightarrow -2^{+}} f(x) =0\)
- \(\displaystyle \displaystyle \lim_{x \rightarrow 2^{-}} f(x)=0\)
1.3.2.4.
Answer.
Many answers are possible; here is one.
1.3.2.5.
Answer.
Many answers are possible; here is one.
1.3.2.6.
Answer.
In general, this is false.
1.3.2.7.
Answer.
False
1.3.2.8.
Answer.
\(\displaystyle\lim_{x \rightarrow -2^-} f(x)=16\)
1.3.2.9.
Answer.
Not enough information to say.
1.3.2.10.
Answer.
\(\displaystyle\lim_{t \rightarrow 0} \sin t=0\)
1.3.2.11.
Answer.
\(\displaystyle\lim_{x \rightarrow 0^+} \log x = -\infty\)
1.3.2.12.
Answer.
\(\displaystyle\lim_{y \rightarrow 3} y^2=9\)
1.3.2.13.
Answer.
\(\displaystyle\lim_{x \rightarrow 0^-} \dfrac{1}{x}=-\infty\)
1.3.2.14.
Answer.
\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x}=\) DNE
1.3.2.15.
Answer.
\(\displaystyle\lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty\)
1.3.2.16.
Answer.
\(\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}=\dfrac{1}{10}\)
1.3.2.17.
Answer.
9
1.4 Calculating Limits with Limit Laws
1.4.2 Exercises
1.4.2.1.
1.4.2.2.
Answer.
There are many possible answers; one is \(f(x)=10(x-3)\text{,}\) \(g(x)=x-3\text{.}\)
1.4.2.3.
Answer.
There are many possible answers; one is \(f(x)=(x-3)^2\) and \(g(x)=x-3\text{.}\) Another is \(f(x)=0\) and \(g(x)=x-3\text{.}\)
1.4.2.4.
Answer.
There are many possible answers; one is \(f(x)=x-3\text{,}\) \(g(x)=(x-3)^3\text{.}\)
1.4.2.5.
Answer.
Any real number; positive infinity; negative infinity; does not exist.
1.4.2.6.
Answer.
0
1.4.2.7.
Answer.
6
1.4.2.8.
Answer.
16
1.4.2.9. (✳).
Answer.
\(4/\cos(3)\)
1.4.2.10. (✳).
Answer.
\(2\)
1.4.2.11. (✳).
Answer.
\(-7/2\)
1.4.2.12. (✳).
Answer.
3
1.4.2.13. (✳).
Answer.
\(-\frac{3}{2}\)
1.4.2.14. (✳).
Answer.
\(\log(2)-1\)
1.4.2.15. (✳).
Answer.
\(\frac{1}{4}\)
1.4.2.16. (✳).
Answer.
\(\dfrac{1}{2}\)
1.4.2.17. (✳).
Answer.
\(5\)
1.4.2.18. (✳).
Answer.
\(-6\)
1.4.2.19.
Answer.
\(-14\)
1.4.2.20. (✳).
Answer.
\(-\frac{1}{3}\)
1.4.2.21. (✳).
Answer.
\(\frac{1}{6} \)
1.4.2.22. (✳).
Answer.
\(\frac{1}{\sqrt{3}}\)
1.4.2.23. (✳).
Answer.
1
1.4.2.24. (✳).
Answer.
12
1.4.2.25.
Answer.
0
1.4.2.26.
Answer.
\(\frac{1}{2}\)
1.4.2.27. (✳).
Answer.
0
1.4.2.28.
Answer.
5
1.4.2.29.
Answer.
\(-\infty\)
1.4.2.30.
Answer.
\(\sqrt{\dfrac{2}{3}}\)
1.4.2.31.
Answer.
DNE
1.4.2.32.
Answer.
\(\infty\)
1.4.2.33.
Answer.
\(x^5-32x+15\)
1.4.2.34.
Answer.
\(0\)
1.4.2.35. (✳).
Answer.
\(0\)
1.4.2.36. (✳).
Answer.
2
1.4.2.37.
Answer.
0
1.4.2.38.
Answer.
\(-\dfrac{32}{9}\)
1.4.2.39.
Answer.
DNE
1.4.2.40.
Answer.
DNE
1.4.2.41.
Answer.
\(-\dfrac{9}{2}\)
1.4.2.42.
Answer.
\(-4\)
1.4.2.43. (✳).
Answer.
\(a=\dfrac{7}{2}\)
1.4.2.44.
Answer.
- \(\displaystyle \displaystyle\lim_{x \rightarrow 0} f(x)=0\)
- \(\displaystyle\lim_{x \rightarrow 0} g(x)=\) DNE
- \(\displaystyle \displaystyle\lim_{x \rightarrow 0} f(x)g(x)=2\)
- \(\displaystyle \displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)}{g(x)}=0\)
- \(\displaystyle \displaystyle\lim_{x \rightarrow 2} f(x)+g(x)=\dfrac{9}{2}\)
- \(\displaystyle \displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)+1}{g(x+1)}=1\)
1.4.2.45.
Answer.
Pictures may vary somewhat; the important points are the values of the function at integer values of \(x\text{,}\) and the vertical asymptotes.
1.4.2.46.
Answer.
1.4.2.47.
Answer.
10
1.4.2.48.
Answer.
1.4.2.48.a DNE , DNE
1.4.2.48.c No: it is only true when both \(\displaystyle\lim_{x \rightarrow a} f(x)\) and \(\displaystyle\lim_{x \rightarrow a} g(x)\) exist.
1.4.2.49.
Answer.
1.4.2.49.a \(\displaystyle\lim_{x \rightarrow 0^-} f(x)=-3\)
1.4.2.49.b \(\displaystyle\lim_{x \rightarrow 0^+} f(x)=3\)
1.4.2.49.c \(\displaystyle\lim_{x \rightarrow 0} f(x)=\)DNE
1.4.2.50.
Answer.
1.4.2.50.a \(\displaystyle\lim_{x \rightarrow -4^-} f(x)=0\)
1.4.2.50.b \(\displaystyle\lim_{x \rightarrow -4^+} f(x)=0\)
1.4.2.50.c \(\displaystyle\lim_{x \rightarrow -4} f(x)=0\)
1.5 Limits at Infinity
1.5.2 Exercises
1.5.2.1.
Answer.
There are many answers: any constant polynomial has this property. One answer is \(f(x)=1\text{.}\)
1.5.2.2.
Answer.
There are many answers: any odd-degree polynomial has this property. One answer is \(f(x)=x\text{.}\)
1.5.2.3.
Answer.
0
1.5.2.4.
Answer.
\(\infty\)
1.5.2.5.
Answer.
\(0\)
1.5.2.6.
Answer.
DNE
1.5.2.7.
Answer.
\(-\infty\)
1.5.2.8.
Answer.
\(\sqrt{3}\)
1.5.2.9. (✳).
Answer.
3
1.5.2.10. (✳).
Answer.
\(-\frac{3}{4}\)
1.5.2.11. (✳).
Answer.
\(-\dfrac{1}{2}\)
1.5.2.12. (✳).
Answer.
\(\frac{1}{2}\)
1.5.2.13. (✳).
Answer.
\(\frac{5}{3}\)
1.5.2.14. (✳).
Answer.
0
1.5.2.15. (✳).
Answer.
\(\frac{4}{7}\)
1.5.2.16.
Answer.
1
1.5.2.17. (✳).
Answer.
0
1.5.2.18.
Answer.
\(-1\)
1.5.2.19.
Answer.
\(1\)
1.5.2.20. (✳).
Answer.
\(-1\)
1.5.2.21. (✳).
Answer.
\(-\frac{3}{2}\)
1.5.2.22. (✳).
Answer.
\(-\frac{5}{3}\)
1.5.2.23.
Answer.
\(-\infty\)
1.5.2.24. (✳).
Answer.
\(\dfrac{5}{2}\)
1.5.2.25.
Answer.
\(\ds\lim_{a \to 0^+}\dfrac{a^2-\frac{1}{a}}{a-1}=\infty\)
1.5.2.26.
Answer.
\(\ds\lim_{x \to 3}\dfrac{2x+8}{\frac{1}{x-3}+\frac{1}{x^2-9}}=0\)
1.5.2.27.
Answer.
No such rational function exists.
1.5.2.28.
Answer.
This is the amount of the substance that will linger long-term. Since it’s nonzero, the substance would be something that would stay in your body. Something like “tattoo ink” is a reasonable answer, while “penicillin” is not.
1.6 Continuity
1.6.4 Exercises
1.6.4.1.
Answer.
Many answers are possible; the tangent function behaves like this.
1.6.4.2.
Answer.
At some time between my birth and now, I was exactly one meter tall.
1.6.4.3.
Answer.
One example is \(f(x) = \left\{ \begin{array}{ll}
0&\mbox{when }0 \leq x \leq 1\\
2&\mbox{when }1 \lt x \leq 2
\end{array}\right.\text{.}\) The IVT only guarantees \(f(c)=1\) for some \(c\) in \([0,2]\) when \(f\) is continuous over \([0,2]\text{.}\) If \(f\) is not continuous, the IVT says nothing.
1.6.4.4.
Answer.
Yes
1.6.4.5.
Answer.
No
1.6.4.6.
Answer.
No
1.6.4.7.
Answer.
True.
1.6.4.8.
Answer.
True.
1.6.4.9.
Answer.
In general, false.
1.6.4.10.
Answer.
\(\ds\lim_{x \to 0^+} h(x)=0\)
1.6.4.11.
Answer.
\(k=0\)
1.6.4.12.
Answer.
Since \(f\) is a polynomial, it is continuous over all real numbers. \(f(0)=1 \lt 12345\) and \(f(12345)=12345^3+12345^2+12345+1 \gt 12345\) (since all terms are positive). So by the IVT, \(f(c)=12345\) for some \(c\) between \(0\) and \(12345\text{.}\)
1.6.4.13. (✳).
Answer.
\((-\infty, -1)\cup (-1,1) \cup (1,+\infty)\)
1.6.4.14. (✳).
Answer.
\((-\infty, -1)\cup (1,+\infty)\)
1.6.4.15. (✳).
Answer.
The function is continuous except at \(x=\pm \pi, \pm 3\pi, \pm 5\pi, \cdots \text{.}\)
1.6.4.16. (✳).
Answer.
\(x \neq n\pi,\) where \(n\) is any integer
1.6.4.17. (✳).
Answer.
\(\pm 2\)
1.6.4.18. (✳).
Answer.
\(c=1\)
1.6.4.19. (✳).
Answer.
\(-1\text{,}\) \(4\)
1.6.4.20. (✳).
Answer.
\(c=1\text{,}\) \(c=-1\)
1.6.4.21.
Answer.
This isn’t the kind of equality that we can just solve; we’ll need a trick, and that trick is the IVT. The general idea is to show that \(\sin x\) is somewhere bigger, and somewhere smaller, than \(x-1\text{.}\) However, since the IVT can only show us that a function is equal to a constant, we need to slightly adjust our language. Showing \(\sin x = x-1\) is equivalent to showing \(\sin x - x + 1 = 0\text{,}\) so let \(f(x)=\sin x - x +1\text{,}\) and let’s show that it has a real root.
First, we need to note that \(f(x)\) is continuous (otherwise we can’t use the IVT). Now, we need to find a value of \(x\) for which it is positive, and for which it’s negative. By checking a few values, we find \(f(0)\) is positive, and \(f(100)\) is negative. So, by the IVT, there exists a value of \(x\) (between \(0\) and \(100\)) for which \(f(x)=0\text{.}\) Therefore, there exists a value of \(x\) for which \(\sin x = x-1\text{.}\)
1.6.4.22. (✳).
Answer.
We let \(f(x)=3^x-x^2\text{.}\) Then \(f(x)\) is a continuous function, since both \(3^x\) and \(x^2\) are continuous for all real numbers.
We want a value \(a\) such that \(f(a) \gt 0\text{.}\) We see that \(a=0\) works since
\begin{equation*}
f(0)=3^0-0=1 \gt 0.
\end{equation*}
We want a value \(b\) such that \(f(b) \lt 0\text{.}\) We see that \(b=-1\) works since
\begin{equation*}
f(-1)=\frac{1}{3}-1 \lt 0.
\end{equation*}
So, because \(f(x)\) is continuous on \((-\infty, \infty)\) and \(f(0) \gt 0\) while \(f(-1) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (-1,0)\) such that \(f(c)=0\text{.}\)
1.6.4.23. (✳).
Answer.
We let \(f(x)=2\tan(x)-x-1\text{.}\) Then \(f(x)\) is a continuous function on the interval \((-\pi/2, \pi/2)\) since \(\tan(x)=\sin(x)/\cos(x)\) is continuous on this interval, while \(x+1\) is a polynomial and therefore continuous for all real numbers.
We find a value \(a\in (-\pi/2,\pi/2)\) such that \(f(a) \lt 0\text{.}\) We observe immediately that \(a=0\) works since
\begin{equation*}
f(0)=2\tan(0)-0-1=0-1=-1 \lt 0.
\end{equation*}
We find a value \(b\in (-\pi/2, \pi/2)\) such that \(f(b) \gt 0\text{.}\) We see that \(b=\pi/4\) works since
\begin{align*}
f(\pi/4)\amp=2\tan(\pi/4) -\pi/4 - 1=2-\pi/4 - 1=1-\pi/4\\
\amp=(4-\pi)/4 \gt 0
\end{align*}
because \(3 \lt \pi \lt 4\text{.}\)
So, because \(f(x)\) is continuous on \([0,\pi/4]\) and \(f(0) \lt 0\) while \(f(\pi/4) \gt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,\pi/4)\) such that \(f(c)=0\text{.}\)
1.6.4.24. (✳).
Answer.
Let \(f(x) = \sqrt{\cos(\pi x)} - \sin(2\pi x) -1/2\text{.}\) This function is continuous provided \(\cos(\pi x)\geq 0\text{.}\) This is true for \(0 \leq x
\leq \frac{1}{2}\text{.}\)
Now \(f\) takes positive values on \([0,1/2]\text{:}\)
\begin{align*}
f(0) &= \sqrt{\cos(0)} - \sin(0) -1/2 = \sqrt{1} -1/2 = 1/2.
\end{align*}
And \(f\) takes negative values on \([0,1/2]\text{:}\)
\begin{align*}
f(1/2) &= \sqrt{\cos(\pi/2)}-\sin(\pi)-1/2 = 0-0-1/2 = -1/2
\end{align*}
(Notice that \(f(1/3)=(\sqrt{2}-\sqrt{3})/2-1/2\) also works)
So, because \(f(x)\) is continuous on \([0,1/2)\) and \(f(0) \gt 0\) while \(f(1/2) \lt 0\text{,}\) then the Intermediate Value Theorem guarantees the existence of a real number \(c\in (0,1/2)\) such that \(f(c)=0\text{.}\)
1.6.4.25. (✳).
Answer.
We let \(f(x)=\dfrac{1}{\cos^2(\pi x)}-x-\dfrac{3}{2}\text{.}\) Then \(f(x)\) is a continuous function on the interval \((-1/2, 1/2)\) since \(\cos x\) is continuous everywhere and non-zero on that interval.
The function \(f\) takes negative values. For example, when \(x=0\text{:}\)
\begin{align*}
f(0) &= \frac{1}{\cos^2(0)} - 0 - \frac{3}{2} = 1-\frac{3}{2} = -\frac{1}{2} \lt 0.
\end{align*}
It also takes positive values, for instance when \(x=1/4\text{:}\)
\begin{align*}
f(1/4) &= \frac{1}{(\cos \pi/4)^2} - \frac{1}{4} - \frac{3}{2}\\
&= \frac{1}{1/2} - \frac{1+6}{4}\\
&= 2 - 7/4 = 1/4 \gt 0.
\end{align*}
By the IVT there is \(c\text{,}\) \(0 \lt c \lt 1/4\) such that \(f(c)=0\text{,}\) in which case
\begin{gather*}
\dfrac{1}{(\cos\pi c)^2} = c+\dfrac{3}{2}.
\end{gather*}
1.6.4.26.
Answer.
\([0,1]\) is the easiest answer to find. Also acceptable are \([-2,-1]\) and \([14,15]\text{.}\)
1.6.4.27.
Answer.
1.91
1.6.4.28.
Answer.
- If \(f(a)=g(a)\text{,}\) or \(f(b)=g(b)\text{,}\) then we simply take \(c=a\) or \(c=b\text{.}\)
- Suppose \(f(a) \neq g(a)\) and \(f(b) \neq g(b)\text{.}\) Then \(f(a) \lt g(a)\) and \(g(b) \lt f(b)\text{,}\) so if we define \(h(x)=f(x)-g(x)\text{,}\) then \(h(a) \lt 0\) and \(h(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\text{,}\) also \(h\) is continuous over \([a,b]\text{.}\) So, by the Intermediate Value Theorem, there exists some \(c \in (a,b)\) with \(h(c)=0\text{;}\) that is, \(f(c)=g(c)\text{.}\)
2 Derivatives
2.1 Revisiting Tangent Lines
2.1.2 Exercises
2.1.2.1.
Answer.
If \(Q\) is to the left of the \(y\) axis, the secant line has positive slope; if \(Q\) is to the right of the \(y\) axis, the secant line has negative slope.
2.1.2.2.
2.1.2.3.
Answer.
\(\{(a), (c), (e)\}, \{(b),(f)\}, \{(d)\}\)
2.1.2.4.
Answer.
Something like \(1.5\text{.}\) A reasonable answer would be between 1 and 2.
2.1.2.5.
Answer.
There is only one tangent line to \(f(x)\) at \(P\) (shown in blue), but there are infinitely many choices of \(Q\) and \(R\) (one possibility shown in red).
2.1.2.6.
Answer.
2.2 Definition of the Derivative
2.2.4 Exercises
2.2.4.1.
Answer.
(a), (d)
2.2.4.2.
Answer.
(e)
2.2.4.3.
Answer.
(b)
2.2.4.4. (✳).
Answer.
By definition, \(f(x) = x^3\) is differentiable at \(x = 0\) if the limit
\begin{equation*}
\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}
=\lim_{h\rightarrow 0}\frac{h^3-0}{h}
\end{equation*}
exists.
2.2.4.5.
Answer.
\(x=-1\) and \(x=3\)
2.2.4.6.
Answer.
True. (Contrast to Question 2.2.4.7.)
2.2.4.7.
Answer.
In general, false. (Contrast to Question 2.2.4.6.)
2.2.4.8.
Answer.
metres per second
2.2.4.9.
Answer.
\(y-6=3(x-1)\text{,}\) or \(y=3x +3\)
2.2.4.10.
Answer.
\(\dfrac{-1}{x^2}\)
2.2.4.11. (✳).
Answer.
By definition
\begin{equation*}
f'(0)=\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}
=\lim_{h\rightarrow 0}\frac{h|h|}{h}
=\lim_{h\rightarrow 0}|h|=0
\end{equation*}
In particular, the limit exists, so the derivative exists (and is equal to zero).
2.2.4.12. (✳).
Answer.
\(\dfrac{-2}{(x+1)^2}\)
2.2.4.13. (✳).
Answer.
\(\dfrac{-2x}{[x^2+3]^2}\)
2.2.4.14.
Answer.
1
2.2.4.15. (✳).
Answer.
\(f'(x)=-\dfrac{2}{x^3}\)
2.2.4.16. (✳).
Answer.
\(a=4\text{,}\) \(b=-4\)
2.2.4.17. (✳).
Answer.
\(f'(x)=\dfrac{1}{2\sqrt{1+x}}\) when \(x \gt -1\text{;}\) \(f'(x)\) does not exist when \(x \leq -1\text{.}\)
2.2.4.18.
Answer.
\(v(t)=4t^3-2t\)
2.2.4.19. (✳).
Answer.
No, it does not.
2.2.4.20. (✳).
Answer.
No, it does not.
2.2.4.21. (✳).
Answer.
Yes, it is.
2.2.4.22. (✳).
Answer.
Yes, it is.
2.2.4.23.
Answer.
Many answers are possible; here is one.
2.2.4.24.
Answer.
\begin{align*}
p'(x) &= \lim_{h \rightarrow 0} \frac{p(x+h)-p(x)}{h}\\
&= \lim_{h \rightarrow 0} \frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\\
&= \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)+g(x+h)-g(x)}{h}\\
&= \lim_{h \rightarrow 0}\left[\frac{f(x+h)-f(x)}{h}+
\frac{g(x+h)-g(x)}{h}\right]\\
(*)&= \left[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right]+ \left[\lim_{h \rightarrow 0}
\frac{g(x+h)-g(x)}{h}\right]\\
&= f'(x)+g'(x)
\end{align*}
At step (\(*\)), we use the limit law that \(\displaystyle\lim_{x \rightarrow a}\left[ F(x)+G(x)\right] = \displaystyle\lim_{x \rightarrow a} F(x)+\displaystyle\lim_{x \rightarrow a}G(x)\text{,}\) as long as \(\displaystyle\lim_{x \rightarrow a} F(x)\) and \(\displaystyle\lim_{x \rightarrow a}G(x)\) exist. Because the problem states that \(f'(x)\) and \(g'(x)\) exist, we know that \(\displaystyle\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) and \(\displaystyle\lim_{h \rightarrow 0}
\frac{g(x+h)-g(x)}{h}\) exist, so our work is valid.
2.2.4.25.
Answer.
2.2.4.26. (✳).
Answer.
\(y=6x-9\) and \(y=-2x-1\)
2.2.4.27. (✳).
Answer.
\(a \gt 1\)
2.3 Interpretations of the Derivative
2.3.3 Exercises
2.3.3.1.
2.3.3.2.
Answer.
Profit per additional widget sold, when \(t\) widgets are being sold. This is called the marginal profit per widget, when \(t\) widgets are being sold.
2.3.3.3.
Answer.
\(T'(d)\) measures how quickly the temperature is changing per unit change of depth, measured in degrees per metre. \(|T'(d)|\) will probably be largest when \(d\) is near zero, unless there are hot springs or other underwater heat sources.
2.3.3.4.
Answer.
Calories per additional gram, when there are \(w\) grams
2.3.3.5.
Answer.
The acceleration of the object.
2.3.3.6.
Answer.
Degrees Celsius temperature change per joule of heat added. (This is closely related to heat capacity and to specific heat — there’s a nice explanation of this on Wikipedia.)
2.3.3.7.
Answer.
Number of bacteria added per degree. That is: the number of extra bacteria (possibly negative) that will exist in the population by raising the temperature by one degree.
2.3.3.8.
Answer.
\(360R'(t)\)
2.3.3.9.
Answer.
If \(P'(t)\) is positive, your sample is below the ideal temperature, and if \(P'(t)\) is negative, your sample is above the ideal temperature. If \(P'(t) = 0\text{,}\) you don’t know whether the sample is exactly at the ideal temperature, or way above or below it with no living bacteria.
2.4 Arithmetic of Derivatives - a Differentiation Toolbox
2.4.2 Exercises
2.4.2.1.
Answer.
True
2.4.2.2.
Answer.
False, in general.
2.4.2.3.
Answer.
True
2.4.2.4.
Answer.
If you’re creative, you can find lots of ways to differentiate!
- Constant multiple: \(g'(x)=3f'(x)\text{.}\)
- Product rule: \(g'(x) = \diff{}{x}\{3\}f(x)+3f'(x)=0f(x)+3f'(x)=3f'(x)\text{.}\)
- Sum rule: \(g'(x)=\diff{}{x}\{f(x)+f(x)+f(x)\}=f'(x)+f'(x)+f'(x)=3f'(x)\text{.}\)
- Quotient rule: \(g'(x)=\diff{}{x}\left\{\frac{f(x)}{\frac{1}{3}}\right\}= \frac{\frac{1}{3}f'(x)-f(x)(0)}{\frac{1}{9}}=\frac{\frac{1}{3}f'(x)}{\frac{1}{9}}=9\left(\frac{1}{3}\right)f'(x)=3f'(x)\text{.}\)
All rules give \(g'(x)=3f'(x)\text{.}\)
2.4.2.5.
Answer.
\(f'(x)=6x+\frac{2}{\sqrt{x}}\)
2.4.2.6.
Answer.
\(-36x+24\sqrt{x}+\frac{20}{\sqrt{x}}-45\)
2.4.2.7. (✳).
Answer.
\(y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right)\text{,}\) or \(y= \tfrac{3}{4} x - \tfrac{1}{4}\)
2.4.2.8. (✳).
2.4.2.9. (✳).
Answer.
\(\dfrac{1}{{(x+1/2)}^2}\text{,}\) or \(\dfrac{4}{(2x+1)^2}\)
2.4.2.10.
Answer.
\(-72\)
2.4.2.11.
Answer.
\(y-\frac{1}{2}=-\frac{1}{8}(x-1)\text{,}\) or \(y=-\tfrac{1}{8}x +\tfrac{5}{8}\)
2.4.2.12.
Answer.
\(b'(t)-d'(t)\)
2.4.2.13. (✳).
Answer.
{\((1,3),\ (3,27)\)}
2.4.2.14. (✳).
Answer.
\(\dfrac{1}{2\sqrt{100180}}\)
2.4.2.15.
Answer.
\(20t+7\) square metres per second.
2.4.2.16.
Answer.
0
2.4.2.17.
Answer.
First expression, \(f(x)=\dfrac{g(x)}{h(x)}\text{:}\)
\begin{align*}
f'(x)&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)}
\end{align*}
Second expresson, \(f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}\text{:}\)
\begin{align*}
\amp f'(x)=\left(\frac{k(x)g'(x)-g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)\\
\amp\hskip1in+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)-k(x)h'(x)}{h^2(x)}\right)\\
&=\frac{k(x)g'(x)-g(x)k'(x)}{k(x)h(x)}+
\frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)}{k(x)h^2(x)}+
\frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)+g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)k(x)g'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\
&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)}
\end{align*}
and this is exactly what we got from differentiating the first expression.
2.6 Using the Arithmetic of Derivatives – Examples
2.6.2 Exercises
2.6.2.1.
Answer.
In the quotient rule, there is a minus, not a plus. Also, \(2(x+1)+2x\) is not the same as \(2(x+1)\text{.}\)
The correct version is:
\begin{align*}
f(x)&=\frac{2x}{x+1}\\
f'(x)&=\frac{2(x+1)\textcolor{red}{-}2x}{(x+1)^2}\\
&=\frac{2}{(x+1)^2}
\end{align*}
2.6.2.2.
Answer.
False. Lemma 2.6.9 does not apply. See the solution.
2.6.2.3.
Answer.
\(4x(x^2+2)(x^2+3)\)
2.6.2.4.
Answer.
\(12t^3+15t^2+\frac{1}{t^2}\)
2.6.2.5.
Answer.
\(x'(y)=8y^3+2y\)
2.6.2.6.
Answer.
\(T'(x)=\dfrac{(x^2+3)\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{x}+1)(2x)}{(x^2+3)^2}\)
2.6.2.7. (✳).
Answer.
\(\dfrac{21-4x-7x^2}{(x^2+3)^2}\)
2.6.2.8.
Answer.
7
2.6.2.9.
Answer.
\(\dfrac{3x^4+30x^3-2x-5}{(x^2+5x)^2}\)
2.6.2.10. (✳).
Answer.
\(\dfrac{-3x^2+12x+5}{(2-x)^2}\)
2.6.2.11. (✳).
Answer.
\(\dfrac{-22x}{(3x^2+5)^2}\)
2.6.2.12. (✳).
Answer.
\(\dfrac{4x^3+12x^2-1}{(x+2)^2}\)
2.6.2.13. (✳).
Answer.
The derivative of the function is
\begin{align*}
\frac{(1-x^2)\cdot\frac{1}{2\sqrt{x}} - \sqrt{x} \cdot (-2x)}{(1-x^2)^2}
&= \frac{(1-x^2) - 2x \cdot (-2x)}{2\sqrt{x}(1-x^2)^2}
\end{align*}
The derivative is undefined if either \(x \lt 0\) or \(x = 0,\pm 1\) (since the square-root is undefined for \(x \lt 0\) and the denominator is zero when \(x=0,1,-1\text{.}\) Putting this together — the derivative exists for \(x \gt 0, x\neq 1\text{.}\)
2.6.2.14.
Answer.
\(\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+
\left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right)\)
2.6.2.15.
Answer.
\(f'(x)=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right)\)
2.6.2.16.
Answer.
\(x=-5\) and \(x=1\)
2.6.2.17. (✳).
Answer.
\(y=x-\dfrac{1}{4}\)
2.6.2.18.
Answer.
\(y=4x-4\) and \(y=-2x-1\)
2.6.2.19. (✳).
Answer.
\(2015\cdot 2^{2014}\)
2.7 Derivatives of Exponential Functions
2.7.3 Exercises
2.7.3.1.
Answer.
A-\((a)\) and \((d)\text{,}\) B-\((e)\text{,}\) C-\((c)\text{,}\) D-\((b)\)
2.7.3.2.
Answer.
\((b),\;\;(d),\;\;(e)\)
2.7.3.3.
Answer.
False
2.7.3.4.
Answer.
increasing
2.7.3.5.
Answer.
\(\dfrac{(x-1)e^x}{2x^2}\)
2.7.3.6.
Answer.
\(2e^{2x}\)
2.7.3.7.
Answer.
\(e^{a+x}\)
2.7.3.8.
Answer.
\(x \gt -1\)
2.7.3.9.
Answer.
\(-e^{-x}\)
2.7.3.10.
Answer.
\(2e^{2x}\)
2.7.3.11.
Answer.
When \(t\) is in the interval \((-2,0)\text{.}\)
2.7.3.12.
Answer.
\(g'(x)=[f(x)+f'(x)]e^x\)
2.7.3.13.
Answer.
(b) and (d)
2.7.3.14. (✳).
Answer.
\(a=b=\dfrac{e}{2}\)
2.8 Derivatives of Trigonometric Functions
2.8.8 Exercises
2.8.8.1.
Answer.
The graph \(f(x)=\sin x\) has horizontal tangent lines precisely at those points where \(\cos x=0\text{.}\)
2.8.8.2.
Answer.
The graph \(f(x)=\sin x\) has maximum slope at those points where \(\cos x\) has a maximum. That is, where \(\cos x = 1\text{.}\)
2.8.8.3.
Answer.
\(f'(x)=\cos x - \sin x + \sec^2 x\)
2.8.8.4.
Answer.
\(x=\frac{\pi}{4}+\pi n\text{,}\) for any integer \(n\text{.}\)
2.8.8.5.
Answer.
0
2.8.8.6.
Answer.
\(f'(x)=2(\cos^2 x - \sin ^2 x)\)
2.8.8.7.
Answer.
\(f'(x)=e^x(\cot x - \csc^2 x)\)
2.8.8.8.
Answer.
\(f'(x)=\dfrac{2+3 \sec x + 2 \sin x -2\tan x \sec x+3\sin x \tan x }{(\cos x + \tan x)^2}\)
2.8.8.9.
Answer.
\(f'(x)=\dfrac{5\sec x \tan x - 5 \sec x - 1}{e^x}\)
2.8.8.10.
Answer.
\(f'(x)=(e^x+\cot x)(30x^5+\csc x \cot x)+(e^x-\csc^2x)(5x^6-\csc x)\)
2.8.8.11.
Answer.
\(-\sin(\theta)\)
2.8.8.12.
Answer.
\(f'(x)=-\cos x - \sin x\)
2.8.8.13.
Answer.
\(\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2+1\)
2.8.8.14. (✳).
Answer.
\(a=0\text{,}\) \(b=1\text{.}\)
2.8.8.15. (✳).
Answer.
\(y - \pi = 1\cdot (x-\pi/2)\)
2.8.8.16. (✳).
Answer.
\(-\sin(2015)\)
2.8.8.17. (✳).
Answer.
\(-\sqrt{3}/2\)
2.8.8.18. (✳).
Answer.
\(-1\)
2.8.8.19.
Answer.
\begin{align*}
\tan \theta &= \dfrac{\sin \theta}{\cos \theta}\\
\end{align*}
So, using the quotient rule,
\begin{align*} \diff{}{\theta}\{\tan \theta\}&=\frac{\cos\theta\cos\theta-\sin\theta(-\sin\theta)}{\cos^2\theta} =\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}\\ &=\left(\frac{1}{\cos \theta}\right)^2=\sec^2\theta \end{align*}2.8.8.20. (✳).
Answer.
\(a=-\frac{2}{3}\text{,}\) \(b=2\)
2.8.8.21. (✳).
Answer.
All values of \(x\) except \(x=\frac{\pi}{2}+n\pi\text{,}\) for any integer \(n\text{.}\)
2.8.8.22. (✳).
Answer.
The function is differentiable whenever \(x^2+x-6\ne 0\) since the derivative equals
\begin{gather*}
\frac{10\cos(x)\cdot (x^2+x-6)-10\sin(x)\cdot (2x+1)}{(x^2+x-6)^2},
\end{gather*}
which is well-defined unless \(x^2+x-6=0\text{.}\) We solve \(x^2+x-6=(x-2)(x+3)=0,\) and get \(x=2\) and \(x=-3\text{.}\) So, the function is differentiable for all real values \(x\) except for \(x=2\) and for \(x=-3\text{.}\)
2.8.8.23. (✳).
Answer.
The function is differentiable whenever \(\sin(x)\ne 0\) since the derivative equals
\begin{gather*}
\frac{\sin(x)\cdot (2x+6) - \cos(x)\cdot (x^2+6x+5)}{(\sin x)^2},
\end{gather*}
which is well-defined unless \(\sin x = 0\text{.}\) This happens when \(x\) is an integer multiple of \(\pi\text{.}\) So, the function is differentiable for all real values \(x\) except \(x=n\pi\text{,}\) where \(n\) is any integer.
2.8.8.24. (✳).
Answer.
\(y - 1 = 2\cdot (x-\pi/4)\)
2.8.8.25. (✳).
Answer.
\(y=2x+2\)
2.8.8.26.
Answer.
\(x = \frac{3\pi}{4}+n\pi\) for any integer \(n\text{.}\)
2.8.8.27.
Answer.
\(f'(0)=0\)
2.8.8.28. (✳).
Answer.
\(h'(x)=\left\{\begin{array}{rl}
\cos x&x \gt 0\\
-\cos x&x \lt 0
\end{array}\right.\) It exists for all \(x \neq 0\text{.}\)
2.8.8.29. (✳).
Answer.
2.8.8.30. (✳).
Answer.
\(2\)
2.9 One More Tool – the Chain Rule
2.9.4 Exercises
2.9.4.3.
Answer.
\(-5\sin(5x+3)\)
2.9.4.4.
Answer.
\(10x(x^2+2)^4\)
2.9.4.5.
Answer.
\(17(4k^4+2k^2+1)^{16}\cdot(16k^3+4k)\)
2.9.4.6.
Answer.
\(\frac{-2x}{({x^2-1})\sqrt{x^4-1}}\)
2.9.4.7.
Answer.
\(-e^{\cos(x^2)}\cdot \sin(x^2)\cdot 2x\)
2.9.4.8. (✳).
Answer.
\(-4\)
2.9.4.9. (✳).
Answer.
\([\cos x -x\sin x]e^{x\cos(x)}\)
2.9.4.10. (✳).
Answer.
\([2x-\sin x]e^{x^2+\cos(x)}\)
2.9.4.11. (✳).
Answer.
\(\frac{3}{2\sqrt{x-1}\sqrt{x+2}^3}\)
2.9.4.12. (✳).
Answer.
\(f'(x)= -\dfrac{2}{x^3}+\dfrac{x}{\sqrt{x^2-1}}\) is defined for \(x\) in \((-\infty,-1) \cup (1,\infty)\text{.}\)
2.9.4.13. (✳).
Answer.
\(f'(x)=\dfrac{(1+x^2)(5\cos 5x)-(\sin 5x)(2x)}{{(1+x^2)}^2}\)
2.9.4.14.
Answer.
\(2e^{2x+7}\sec(e^{2x+7})\tan(e^{2x+7}) \)
2.9.4.15.
Answer.
\(y=1\)
2.9.4.16.
Answer.
\(t=\frac{2}{3}\) and \(t=4\)
2.9.4.17.
Answer.
\(2e\sec^2(e)\)
2.9.4.18. (✳).
Answer.
\(y'=4e^{4x}\tan x+e^{4x}\sec^2 x\)
2.9.4.19. (✳).
Answer.
\(\dfrac{3}{{(1+e^{3})}^2}\)
2.9.4.20. (✳).
Answer.
\(2 \sin(x) \cdot \cos(x) \cdot e^{\sin^2(x)}\)
2.9.4.21. (✳).
Answer.
\(\cos\left(e^{5x}\right)\cdot e^{5x}\cdot 5\)
2.9.4.22. (✳).
Answer.
\(-e^{\cos(x^2)}\cdot \sin(x^2) \cdot 2x\)
2.9.4.23. (✳).
Answer.
\(y'=-\sin\big(x^2+\sqrt{x^2+1}\big)\left(2x+\dfrac{x}{\sqrt{x^2+1}}\right)\)
2.9.4.24. (✳).
Answer.
\(y'=2x\cos^2x-2(1+x^2) \sin x\cos x\)
2.9.4.25. (✳).
Answer.
\(y'=\dfrac{e^{3x}(3x^2-2x+3)}{(1+x^2)^2}\)
2.9.4.26. (✳).
Answer.
\(-40\)
2.9.4.27. (✳).
Answer.
\((1,1)\) and \((-1,-1)\text{.}\)
2.9.4.28.
Answer.
Always
2.9.4.29.
Answer.
\(e^x\sec^3(5x-7)(1+15\tan(5x-7))\)
2.9.4.30. (✳).
Answer.
\(e^{2x} \cos 4x + 2x e^{2x} \cos 4x -4 x e^{2x} \sin 4x\)
2.9.4.31.
Answer.
\(t=\dfrac{\pi}{4}\)
2.9.4.32. (✳).
Answer.
Let \(f(x)=e^{x+x^2}\) and \(g(x)=1+x\text{.}\) Then \(f(0)=g(0)=1\text{.}\)
\(f'(x)=(1+2x)e^{x+x^2}\) and \(g'(x)=1\text{.}\) When \(x \gt 0\text{,}\)
\begin{equation*}
f'(x)=(1+2x)e^{x+x^2} \gt 1\cdot e^{x+x^2}=e^{x+x^2} \gt e^{0+0^2}=1=g'(x).
\end{equation*}
Since \(f(0)=g(0)\text{,}\) and \(f'(x) \gt g'(x)\) for all \(x \gt 0\text{,}\) that means \(f\) and \(g\) start at the same place, but \(f\) always grows faster. Therefore, \(f(x) \gt g(x)\) for all \(x \gt 0\text{.}\)
2.9.4.33.
Answer.
\(\cos(2x)=\cos^2x-\sin^2x\)
2.9.4.34.
Answer.
\begin{align*}
f'(x)&=
\frac{1}{3}\left(
\dfrac{ \sqrt{x^3-9} \tan x }{e^{\csc x^2}}
\right)^{\frac{2}{3}}\cdot\\
&\hskip0.3in\frac{e^{\csc x^2} \Big( -2x\sqrt{x^3\!-\!9}\tan x\frac{\cos(x^2)}{\sin^2(x^2)}
-\frac{3x^2\tan x}{2\sqrt{x^3-9}}-\sqrt{x^3\!-\!9}\sec^2 x\Big)}
{(\tan^2 x)(x^3-9) }
\end{align*}
2.9.4.35.
Answer.
The particle traces the curve \(y=1-x^2\) restricted to domain \([-1,1]\text{.}\) At \(t=0\text{,}\) the particle is at the top of the curve, \((1,0)\text{.}\) Then it moves to the right, and goes back and forth along the curve, repeating its path every \(2\pi\) units of time.
2.9.4.35.b \(\sqrt{3}\)
2.10 The Natural Logarithm
2.10.3 Exercises
2.10.3.1.
Answer.
Ten speakers: 13 dB. One hundred speakers: 23 dB.
2.10.3.2.
Answer.
\(20\log2 \approx 14\) years
2.10.3.3.
Answer.
(b)
2.10.3.4.
Answer.
\(f'(x)=\dfrac{1}{x}\)
2.10.3.5.
Answer.
\(f'(x)=\dfrac{2}{x}\)
2.10.3.6.
Answer.
\(f'(x)=\dfrac{2x+1}{x^2+x}\)
2.10.3.7.
Answer.
\(f'(x) = \dfrac{1}{x\log10}\)
2.10.3.8. (✳).
Answer.
\(y'=\dfrac{1-3\log x}{x^4}\)
2.10.3.9.
Answer.
\(\ds\diff{}{\theta} \log(\sec \theta) = \tan \theta\)
2.10.3.10.
Answer.
\(f'(x)=\dfrac{-e^{\cos(\log x)}\sin(\log x)}{x}\)
2.10.3.11. (✳).
Answer.
\(y'=\dfrac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}\)
2.10.3.12. (✳).
Answer.
\(\dfrac{\tan x}{2\sqrt{-\log(\cos x)}}\)
2.10.3.13. (✳).
Answer.
\(\dfrac{\sqrt{x^2+4}+x}{x\sqrt{x^2+4}+x^2+4}=\dfrac{1}{\sqrt{x^2+4}}\)
2.10.3.14. (✳).
Answer.
\(g'(x)=\dfrac{2xe^{x^2}\sqrt{1+x^4}+2x^3}{e^{x^2}\sqrt{1+x^4}+1+x^4}\)
2.10.3.15. (✳).
Answer.
\(\dfrac{4}{3}\)
2.10.3.16.
Answer.
\(f'(x)=\dfrac{3x}{x^2+5}-\dfrac{2x^3}{x^4+10}\)
2.10.3.17.
Answer.
\(\dfrac{40}{3}\)
2.10.3.18. (✳).
Answer.
\(g'(x)=\pi^x\log \pi+\pi x^{\pi -1}\)
2.10.3.19.
Answer.
\(f'(x) = x^x(\log x + 1)\)
2.10.3.20. (✳).
Answer.
\(x^x(\log x+1)+\dfrac{1}{x\log 10}\)
2.10.3.21.
Answer.
\(f'(x)=\dfrac{1}{4}\left(
{\sqrt[4]{\dfrac{(x^4\!+\!12)(x^4\!-\!x^2\!+\!2)}{x^3}}}\right)
\left(\dfrac{4x^3}{x^4\!+\!12}+\dfrac{4x^3-2x}{x^4\!-\!x^2\!+\!2}-\dfrac{3}{x}\right)\)
2.10.3.22.
Answer.
\begin{align*}
f'(x)\amp=(x+1)(x^2+1)^2(x^3+1)^3(x^4+1)^4(x^5+1)^5\\
\amp\hskip1.2in\left[\frac{1}{x+1}+\frac{4x}{x^2+1}
+\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1}\right]
\end{align*}
2.10.3.23.
Answer.
\(\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{x^2+2x+3}-\dfrac{6x^2}{3x^4+4x^3+5}-\dfrac{1}{2(x+1)^2}\right)\)
2.10.3.24. (✳).
Answer.
\(f'(x)=(\cos x)^{\sin x}\left[(\cos x) \log (\cos x) - \sin x \tan x\right]\)
2.10.3.25. (✳).
Answer.
\({\ds\diff{}{x}\left\{(\tan x)^x\right\}}={(\tan x)^x}\left(\log(\tan x) + \dfrac{x}{\sin x \cos x}\right)\)
2.10.3.26. (✳).
Answer.
\(2x(x^2+1)^{x^2+1} (1+\log(x^2+1))\)
2.10.3.27. (✳).
Answer.
\(f'(x)= (x^2+1)^{\sin(x)} \cdot \left( \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1}
\right)\)
2.10.3.28. (✳).
Answer.
\(x^{\cos^3(x)} \cdot \left( -3\cos^2(x)\sin(x) \log(x) + \dfrac{\cos^3(x)}{x} \right)\)
2.10.3.29. (✳).
Answer.
\((3+\sin(x))^{x^2-3}\cdot \left[ 2x\log(3+\sin(x)) + \dfrac{(x^2-3)\cos(x)}{3+\sin(x)}\right]\)
2.10.3.30.
Answer.
\(\ds\diff{}{x}\left\{[f(x)]^{g(x)}\right\}=\left[f(x)\right]^{g(x)}\left[
g'(x)\log(f(x))+ \dfrac{g(x)f'(x)}{f(x)}
\right]\)
2.10.3.31.
Answer.
Let \(g(x):=\log(f(x))\text{.}\) Notice \(g'(x)=\frac{f'(x)}{f(x)}\text{.}\)
In order to show that the two curves have horizontal tangent lines at the same values of \(x\text{,}\) we will show two things: first, that if \(f(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(g(x)\) has a horizontal tangent line at that value of \(x\text{.}\) Second, we will show that if \(g(x)\) has a horizontal tangent line at some value of \(x\text{,}\) then also \(f(x)\) has a horizontal tangent line at that value of \(x\text{.}\)
Suppose \(f(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(f'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}\text{.}\) Since \(f(x_0) \neq 0\text{,}\) \(\frac{f'(x_0)}{f(x_0)}=\frac{0}{f(x_0)}=0\text{,}\) so \(g(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(f\) has a horizontal tangent line, \(g\) has one too.
Now suppose \(g(x)\) has a horizontal tangent line where \(x=x_0\) for some point \(x_0\text{.}\) This means \(g'(x_0)=0\text{.}\) Then \(g'(x_0)=\frac{f'(x_0)}{f(x_0)}=0\text{,}\) so \(f'(x_0)\) exists and is equal to zero. Therefore, \(f(x)\) also has a horizontal tangent line when \(x=x_0\text{.}\) This shows that whenever \(g\) has a horizontal tangent line, \(f\) has one too.
2.11 Implicit Differentiation
2.11.2 Exercises
2.11.2.1.
Answer.
(a) and (b)
2.11.2.2.
Answer.
At \((0,4)\) and \((0,-4)\text{,}\) \(\ds\diff{y}{x}\) is 0; at \((0,0)\text{,}\) \(\ds\diff{y}{x}\) does not exist.
2.11.2.3.
Answer.
(a) no
(b) no \(\ds\diff{y}{x}=-\dfrac{x}{y}\text{.}\) It is not possible to write \(\ds\diff{y}{x}\) as a function of \(x\text{,}\) because (as stated in (b)) one value of \(x\) may give two values of \(\ds\diff{y}{x}\text{.}\) For instance, when \(x=\pi/4\text{,}\) at the point \(\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)\) the circle has slope \(\ds\diff{y}{x}=-1\text{,}\) while at the point \(\left(\dfrac{\pi}{4},\dfrac{-1}{\sqrt{2}}\right)\) the circle has slope \(\ds\diff{y}{x}=1\text{.}\)
2.11.2.4. (✳).
Answer.
\(\ds\diff{y}{x}=-\dfrac{e^x+y}{e^y+x}\)
2.11.2.5. (✳).
Answer.
\(\ds\diff{y}{x} = \dfrac{y^2+1}{e^y-2xy}\)
2.11.2.6. (✳).
Answer.
At \((x,y)=(4,1)\text{,}\) \(y' =
-\dfrac{1}{\pi + 1}\text{.}\) At \((x,y)=(-4,1)\text{,}\) \(y' = \dfrac{1}{\pi-1}\text{.}\)
2.11.2.7. (✳).
Answer.
\(-\ds\frac{2x\sin(x^2+y)+3x^2}{4y^3+\sin(x^2+y)}\)
2.11.2.8. (✳).
Answer.
At \((x,y)=(1,0)\text{,}\) \(y' = -6\text{,}\) and at \((x,y)=(-5,0)\text{,}\) \(y' = \frac{6}{25}\text{.}\)
2.11.2.9. (✳).
Answer.
\(\diff{y}{x} = \dfrac{\cos(x+y)-2x}{2y-\cos(x+y)}\)
2.11.2.10. (✳).
Answer.
At \((x,y)=(2,0)\) we have \(y' = -\frac{3}{2}\text{,}\) and at \((x,y)=(-4,0)\) we have \(y' = -\frac{3}{4}\text{.}\)
2.11.2.11.
Answer.
\(\left(\dfrac{\sqrt{3}}{2},\dfrac{-1}{2\sqrt{3}}\right)\text{,}\) \(\left(\dfrac{-\sqrt{3}}{2},\dfrac{1}{2\sqrt{3}}\right)\)
2.11.2.12. (✳).
Answer.
\(-\dfrac{28}{3}\)
2.11.2.13. (✳).
Answer.
\(\ds\diff{y}{x}=-\dfrac{2xy^2+\sin y}{2x^2y+x\cos y}\)
2.11.2.14. (✳).
Answer.
At \((x,y)=(2,0)\text{,}\) \(y' = -2\text{.}\) At \((x,y)=(-2,0)\text{,}\) \(y' = 2\text{.}\)
2.11.2.15.
Answer.
\(x=0\text{,}\) \(x=1\text{,}\) \(x=-1\)
2.12 Inverse Trigonometric Functions
2.12.2 Exercises
2.12.2.1.
Answer.
(a) \((-\infty,\infty)\)
(b) all integer multiples of \(\pi\)
(c) \([-1,1]\)
2.12.2.2.
Answer.
False
2.12.2.3.
Answer.
2.12.2.4.
Answer.
- If \(|a| \gt 1\text{,}\) there is no point where the curve has horizontal tangent line.
- If \(|a|=1\text{,}\) the curve has a horizontal tangent line where \(x=2\pi n + \dfrac{a\pi}{2}\) for any integer \(n\text{.}\)
- If \(|a| \lt 1\text{,}\) the curve has a horizontal tangent line where \(x=2\pi n+\arcsin(a)\) or \(x=(2 n +1) \pi - \arcsin (a)\) for any integer \(n\text{.}\)
2.12.2.5.
Answer.
Domain: \(x=\pm 1\text{.}\) Not differentiable anywhere.
2.12.2.6.
Answer.
\(f'(x)=\dfrac{1}{\sqrt{9-x^2}}\text{;}\) domain of \(f\) is \([-3,3]\text{.}\)
2.12.2.7.
Answer.
\(f'(t)=\dfrac{-\frac{t^2-1}{\sqrt{1-t^2}}-2t\arccos t}{(t^2-1)^2}\text{,}\) and the domain of \(f(t)\) is \((-1,1)\text{.}\)
2.12.2.8.
Answer.
The domain of \(f(x)\) is all real numbers, and \(f'(x)=\dfrac{-2x}{(x^2+2)\sqrt{x^4+4x^2+3}}\text{.}\)
2.12.2.9.
Answer.
\(f'(x)=\dfrac{1}{a^2+x^2}\) and the domain of \(f(x)\) is all real numbers.
2.12.2.10.
Answer.
\(f'(x)=\arcsin x\text{,}\) and the domain of \(f(x)\) is \([-1,1]\text{.}\)
2.12.2.11.
Answer.
\(x=0\)
2.12.2.12.
Answer.
\(\ds\diff{}{x}\{\arcsin x + \arccos x\}=0\)
2.12.2.13. (✳).
Answer.
\(y'=\dfrac{-1}{x^2\sqrt{1-\frac{1}{x^2}}}\)
2.12.2.14. (✳).
Answer.
\(y'=\dfrac{-1}{1+x^2}\)
2.12.2.15. (✳).
Answer.
\(2x\arctan x+1\)
2.12.2.16.
Answer.
Let \(\theta = \arctan x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\tan \theta = x\text{.}\) In particular, the ratio of the opposite side to the adjacent side is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the hypotenuse came from the Pythagorean Theorem. Now,
\begin{equation*}
\sin\left(\arctan x\right) = \sin \theta = \frac{\mbox{opp}}{\mbox{hyp}} = \frac{x}{\sqrt{x^2+1}}
\end{equation*}
From here, we differentiate using the quotient rule:
\begin{align*}
\diff{}{x}\left\{\frac{x}{\sqrt{x^2+1}}
\right\}&=
\frac{\sqrt{x^2+1}-x\frac{2x}{2\sqrt{x^2+1}}}{x^2+1}\\
&=\left(\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}\right)\cdot\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\\
&=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\
&=\frac{1}{(x^2+1)^{3/2}}=(x^2+1)^{-3/2}
\end{align*}
2.12.2.17.
Answer.
Let \(\theta = \arcsin x\text{.}\) Then \(\theta\) is the angle of a right triangle that gives \(\sin \theta = x\text{.}\) In particular, the ratio of the opposite side to the hypotenuse is \(x\text{.}\) So, we have a triangle that looks like this:
where the length of the adjacent side came from the Pythagorean Theorem. Now,
\begin{equation*}
\cot\left(\arcsin x\right) = \cot \theta = \frac{\mbox{adj}}{\mbox{opp}} = \frac{\sqrt{1-x^2}}{x}
\end{equation*}
From here, we differentiate using the quotient rule:
\begin{align*}
\diff{}{x}\left\{\frac{\sqrt{1-x^2}}{x}
\right\}&=
\frac{x\frac{-2x}{2\sqrt{1-x^2}}-\sqrt{1-x^2}}{x^2}\\
&=\frac{-x^2-(1-x^2)}{x^2\sqrt{1-x^2}}\\
&=\frac{-1}{x^2\sqrt{1-x^2}}
\end{align*}
2.12.2.18. (✳).
Answer.
\((x,y)=\pm\big(\frac{\sqrt{3}}{2},\frac{\pi}{3}\big)\)
2.12.2.19.
Answer.
\(x=\dfrac{(2n+1)\pi}{2}\) for any integer \(n\)
2.12.2.20. (✳).
Answer.
\(g'(y)=\dfrac{1}{1-\sin g(y)}\)
2.12.2.21. (✳).
Answer.
\(\dfrac{1}{2}\)
2.12.2.22. (✳).
Answer.
\(\dfrac{1}{e+1}\)
2.12.2.23.
Answer.
\(f'(x)=[\sin x +2]^{\arcsec x}\left(\dfrac{\log[\sin x +2]}{|x|\sqrt{x^2-1}}+ \dfrac{\arcsec x \cdot\cos x}{\sin x +2}\right)\text{.}\) The domain of \(f(x)\) is \(|x|\ge 1\text{.}\)
2.12.2.24.
Answer.
The function \(\dfrac{1}{\sqrt{x^2-1}}\) exists only for those values of \(x\) with \(x^2-1 \gt 0\text{:}\) that is, the domain of \(\dfrac{1}{\sqrt{x^2-1}}\) is \(|x| \gt 1\text{.}\) However, the domain of arcsine is \(|x| \leq 1\text{.}\) So, there is not one single value of \(x\) where \(\arcsin x\) and \(\dfrac{1}{\sqrt{x^2-1}}\) are both defined.
If the derivative of \(\arcsin(x)\) were given by \(\dfrac{1}{\sqrt{x^2-1}}\text{,}\) then the derivative of \(\arcsin(x)\) would not exist anywhere, so we would probably just write “derivative does not exist,” instead of making up a function with a mismatched domain. Also, the function \(f(x)=\arcsin(x)\) is a smooth curve--its derivative exists at every point strictly inside its domain. (Remember not all curves are like this: for instance, \(g(x)=|x|\) does not have a derivative at \(x=0\text{,}\) but \(x=0\) is strictly inside its domain.) So, it’s a pretty good bet that the derivative of arcsine is not \(\dfrac{1}{\sqrt{x^2-1}}\text{.}\)
2.12.2.25.
Answer.
\(\dfrac{1}{2}\)
2.12.2.26.
Answer.
\(f^{-1}(7)=-\dfrac{25}{4}\)
2.12.2.27.
Answer.
\(f(0)=-7\)
2.12.2.28.
Answer.
\(y'=\dfrac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}}\text{,}\) or equivalently, \(y'=\dfrac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)}\)
2.13 The Mean Value Theorem
2.13.5 Exercises
2.13.5.1.
Answer.
The caribou spent at least about 71 and a half hours travelling during its migration (probably much more) in one year.
2.13.5.2.
Answer.
At some point during the day, the crane was travelling at exactly 10 kph.
2.13.5.3.
Answer.
One possible answer:
Another possible answer:
2.13.5.4.
Answer.
One possible answer: \(f(x) = \left\{\begin{array}{lr}
0&x \neq 10\\
10&x=10
\end{array}\right.\)
Another answer: \(f(x) = \left\{\begin{array}{lr}
10&x \neq 0\\
0&x=0
\end{array}\right.\)
Yet another answer: \(f(x) = \left\{\begin{array}{ll}
5&x \neq 0, 10\\
10&x= 10\\
0&x=0
\end{array}\right.\)
2.13.5.5.
Answer.
(a) No such function is possible: Rolle’s Theorem guarantees \(f'(c)=0\) for at least one point \(c \in (1,2)\text{.}\)
For the other functions, examples are below, but many answers are possible.
2.13.5.6.
Answer.
The function \(f(x)\) is continuous over all real numbers, but it is only differentiable when \(x \neq 0\text{.}\) So, if we want to apply the MVT, our interval must consist of only positive numbers or only negative numbers: the interval \((-4,13)\) is not valid.
It is possible to use the mean value theorem to prove what we want: if \(a=1\) and \(b=144\text{,}\) then \(f(x)\) is differentiable over the interval \((1,144)\) (since 0 is not contained in that interval), and \(f(x)\) is continuous everywhere, so by the mean value theorem there exists some point \(c\) where \(f'(x)=\dfrac{\sqrt{|144|}-\sqrt{|1|}}{144-1}=\dfrac{11}{143}=\dfrac{1}{13}\text{.}\)
That being said, an easier way to prove that a point exists is to simply find it--without using the MVT. When \(x \gt 0\text{,}\) \(f(x)=\sqrt{x}\text{,}\) so \(f'(x)=\dfrac{1}{2\sqrt {x}}\text{.}\) Then \(f'\left(\dfrac{169}{4}\right)=\dfrac{1}{13}\text{.}\)
2.13.5.7. (✳).
Answer.
We note that \(f(0)=f(2\pi)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we conclude that there exists \(c\) in \((0,2\pi)\) such that
\begin{equation*}
f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.
\end{equation*}
2.13.5.8. (✳).
Answer.
We note that \(f(0)=f(1)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in (0,1)\) such that
\begin{equation*}
f'(c)=\frac{f(1)- f(0)}{1 - 0} = 0.
\end{equation*}
2.13.5.9. (✳).
Answer.
We note that \(f(0)=f(2\pi)=\sqrt{3} + \pi^2\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers since \(3+\sin x \gt 0\)), we get that there exists \(c\in (0,2\pi)\) such that
\begin{equation*}
f'(c)=\frac{f(2\pi)- f(0)}{2\pi - 0} = 0.
\end{equation*}
2.13.5.10. (✳).
Answer.
We note that \(f(0)=0\) and \(f(\pi/4)=0\text{.}\) Then using the Mean Value Theorem (note that the function is differentiable for all real numbers), we get that there exists \(c\in
(0,\pi/4)\) such that
\begin{equation*}
f'(c)=\frac{f(\pi/4)- f(0)}{\pi/4 - 0} = 0.
\end{equation*}
2.13.5.11.
Answer.
\(1\)
2.13.5.12.
Answer.
\(2\)
2.13.5.13.
Answer.
\(1\)
2.13.5.14.
Answer.
\(1\)
2.13.5.15. (✳).
Answer.
\begin{equation*}
f'(x)=15x^4-30x^2+15=15\big(x^4-2x^2+1\big)=15\big(x^2-1\big)^2\ge 0
\end{equation*}
The derivative is nonnegative everywhere. The only values of \(x\) for which \(f'(x)=0\) are \(1\) and \(-1\text{,}\) so \(f'(x) \gt 0\) for every \(x\) in \((-1,1)\text{.}\)
2.13.5.15.b If \(f(x)\) has two roots \(a\) and \(b\) in \([-1,1]\text{,}\) then by Rolle’s Theorem, \(f'(c)=0\) for some \(x\) strictly between \(a\) and \(b\text{.}\) But since \(a\) and \(b\) are in \([-1,1]\text{,}\) and \(c\) is between \(a\) and \(b\text{,}\) that means \(c\) is in \((-1,1)\text{;}\) however, we know for every \(c\) in \((-1,1)\text{,}\) \(f'(c) \gt 0\text{,}\) so this can’t happen. Therefore, \(f(x)\) does not have two roots \(a\) and \(b\) in \([-1,1]\text{.}\) This means \(f(x)\) has at most one root in \([-1,1]\text{.}\)
2.13.5.16. (✳).
Answer.
\(\log\left(\dfrac{e^T-1}{T}\right)\)
2.13.5.17.
Answer.
See the solution for the argument that \(\arcsec x=C-\arccsc x\) for some constant \(C\text{.}\)
The constant \(C=\dfrac{\pi}{2}\text{.}\)
2.13.5.18. (✳).
Answer.
Since \(e^{-f(x)}\) is always positive (regardless of the value of \(f(x)\)),
\begin{equation*}
f'(x)=\dfrac{1}{1+e^{-f(x)}} \lt \dfrac{1}{1+0}=1
\end{equation*}
for every \(x\text{.}\)
Since \(f'(x)\) exists for every \(x\text{,}\) we see that \(f\) is differentiable, so the Mean Value Theorem applies. If \(f(100)\) is greater than or equal to 100, then by the Mean Value Theorem, there would have to be some \(c\) between \(0\) and \(100\) such that
\begin{equation*}
f'(c) = \frac{f(100)-f(0)}{100}\geq\frac{100}{100}= 1
\end{equation*}
Since \(f'(x) \leq 1\) for every \(x\text{,}\) there is no value of \(c\) as described. Therefore, it is not possible that \(f(100) \geq 100\text{.}\) So, \(f(100) \lt 100\text{.}\)
2.13.5.19.
Answer.
Domain of \(f^{-1}(x)\text{:}\) \((-\infty,\infty)\) Interval where \(f\) is one--to--one, and range of \(f^{-1}(x)\text{:}\) \((-\infty,\infty)\)
2.13.5.20.
Answer.
One--to--one interval, and range of \(f^{-1}\text{:}\) \(\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]\) Domain of \(f^{-1}\text{:}\) \(\left[-\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right),\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right]\)
2.13.5.21.
Answer.
Define \(h(x)=f(x)-g(x)\text{,}\) and notice \(h(a)=f(a)-g(a) \lt 0\) and \(h(b)=f(b)-g(b) \gt 0\text{.}\) Since \(h\) is the difference of two functions that are continuous over \([a,b]\) and differentiable over \((a,b)\text{,}\) also \(h\) is continuous over \([a,b]\) and differentiable over \((a,b)\text{.}\) So, by the Mean Value Theorem, there exists some \(c \in (a,b)\) with
\begin{equation*}
h'(c)=\frac{h(b)-h(a)}{b-a}
\end{equation*}
Since \((a,b)\) is an interval, \(b \gt a\text{,}\) so the denominator of the above expression is positive; since \(h(b) \gt 0 \gt h(a)\text{,}\) also the numerator of the above expression is positive. So, \(h'(c) \gt 0\) for some \(c \in (a,b)\text{.}\) Since \(h'(c)=f'(c)-g'(c)\text{,}\) we conclude \(f'(c) \gt g'(c)\) for some \(c \in (a,b)\text{.}\)
2.13.5.22.
Answer.
\(3\)
2.13.5.23.
Answer.
\(2\)
2.14 Higher Order Derivatives
2.14.2 Exercises
2.14.2.1.
Answer.
\(e^x\)
2.14.2.2.
Answer.
2.14.2.3.
Answer.
\(\dfrac{3}{15!}\)
2.14.2.4.
Answer.
The derivative \(\ds\diff{y}{x}\) is \(\dfrac{11}{4}\) only at the point \((1,3)\text{:}\) it is not constantly \(\dfrac{11}{4}\text{,}\) so it is wrong to differentiate the constant \(\dfrac{11}{4}\) to find \(\ds\ddiff{2}{y}{x}\text{.}\) Below is a correct solution.
\begin{align*}
-28x+2y+2xy'+2yy'&=0\\
\end{align*}
Plugging in \(x=1\text{,}\) \(y=3\text{:}\)
\begin{align*} -28+6+2y'+6y'&=0\\ y'&=\frac{11}{4} \quad\text{at the point $(1,3)$}\\ \end{align*}Differentiating the equation \(-28x+2y+2xy'+2yy'=0\text{:}\)
\begin{align*} -28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\ 4y'+2(y')^2+2xy''+2yy''&=28\\ \end{align*}At the point \((1,3)\text{,}\) \(y'=\dfrac{11}{4}\text{.}\) Plugging in:
\begin{align*} 4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\ y''&=\frac{15}{64} \end{align*}2.14.2.5.
Answer.
\(f''(x)=\dfrac{1}{x}\)
2.14.2.6.
Answer.
\(\ds\ddiff{2}{}{x}\{\arctan x\}=\frac{-2x}{(1+x^2)^2}\)
2.14.2.7.
Answer.
\(\ds\ddiff{2}{y}{x}=\dfrac{-1}{y^3}\)
2.14.2.8.
Answer.
\(0\)
2.14.2.9.
Answer.
\(\ds\ddiff{3}{}{x}\{\log(5x^2-12)\}=\frac{100x(5x^2+36)}{(5x^2-12)^3}\)
2.14.2.10.
Answer.
speeding up
2.14.2.11.
Answer.
slower
2.14.2.12.
Answer.
\(-4\)
2.14.2.13.
Answer.
(a) true
(b) true
(c) false
2.14.2.14.
Answer.
(ii)
2.14.2.15.
Answer.
\(f^{(n)}=2^x(\log 2)^n\)
2.14.2.16.
Answer.
\(n=4\)
2.14.2.17. (✳).
Answer.
- 2.14.2.17.c \(f\) and \(h\) “start at the same place”, since \(f(0)=h(0)\text{.}\) Also \(f'(0)=h'(0)\text{,}\) and \(f''(x)=(4x^2+4x+3)e^{x+x^2} \gt 3e^{x+x^2} \gt 3=h''(x)\) when \(x \gt 0\text{.}\) Since \(f'(0)=h'(0)\text{,}\) and since \(f'\) grows faster than \(h'\) for positive \(x\text{,}\) we conclude \(f'(x) \gt h'(x)\) for all positive \(x\text{.}\) Now we can conclude that (since \(f(0)=h(0)\) and \(f\) grows faster than \(h\) when \(x \gt 0\)) also \(f(x) \gt h(x)\) for all positive \(x\text{.}\)
2.14.2.18. (✳).
Answer.
- \(\displaystyle y'(1)=\dfrac{4}{13}\)
2.14.2.19.
Answer.
- 2.14.2.19.a \(g''(x)=[f(x)+2f'(x)+f''(x)]e^x\)
- 2.14.2.19.b \(g'''(x)=[f(x)+3f'(x)+3f''(x)+f'''(x)]e^x\)
- 2.14.2.19.c \(g^{(4)}(x)=[f(x)+4f'(x)+6f''(x)+4f'''(x)+f^{(4)}(x)]e^x\)
2.14.2.20.
Answer.
\(m+n\)
2.14.2.21.
Answer.
\(2\)
2.14.2.22. (✳).
Answer.
2.14.2.22.a In order to make \(f(x)\) a little more tractable, let’s change the format. Since \(|x|=\left\{\begin{array}{rl}
x&x \geq 0\\
-x&x \lt 0
\end{array}\right.\text{,}\) then:
\begin{equation*}
f(x)=\left\{\begin{array}{rl}
-x^2&x \lt 0\\
x^2&x\ge 0.\end{array}
\right.
\end{equation*}
Now, we turn to the definition of the derivative to figure out whether \(f'(0)\) exists.
\begin{align*}
f'(0)&=\lim_{h \to 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(h)-0}{h} =\lim_{h \to 0}\frac{f(h)}{h}\qquad\mbox{if it exists.}\\
\end{align*}
Since \(f\) looks different to the left and right of 0, in order to evaluate this limit, we look at the corresponding one-sided limits. Note that when \(h\) approaches 0 from the right, \(h \gt 0\) so \(f(h)=h^2\text{.}\) By contrast, when \(h\) approaches 0 from the left, \(h \lt 0\) so \(f(h)=-h^2\text{.}\)
\begin{align*} & \lim_{h \to 0^+} \frac{f(h)}{h}=\lim_{h \to 0^+}\frac{h^2}{h}=\lim_{h \to 0^+}h=0\\ & \lim_{h \to 0^-} \frac{f(h)}{h}=\lim_{h \to 0^-}\frac{-h^2}{h}=\lim_{h \to 0^-}-h=0\\ \end{align*}Since both one-sided limits exist and are equal to 0,
\begin{align*} & \lim_{h \to 0} \frac{f(0+h)-f(0)}{h}=0 \end{align*}and so \(f\) is differentiable at \(x=0\) and \(f'(0)=0\text{.}\)
\begin{equation*}
f(x)=\left\{\begin{array}{rl}
-x^2&x \lt 0\\
x^2&x\ge 0.\end{array}
\right.
\end{equation*}
So,
\begin{equation*}
f'(x)=\left\{\begin{array}{rl}
-2x&x \lt 0\\
2x&x\ge 0.\end{array}
\right.
\end{equation*}
Then, we know the second derivative of \(f\) everywhere except at \(x=0\text{:}\)
\begin{equation*}
f''(x)=\left\{\begin{array}{cc}
-2&x \lt 0\\
??&x=0\\
2&x \gt 0.\end{array}
\right.
\end{equation*}
So, whenever \(x \neq 0\text{,}\) \(f''(x)\) exists. To investigate the differentiability of \(f'(x)\) when \(x=0\text{,}\) again we turn to the definition of a derivative. If
\begin{align*}
&\lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}\\
\end{align*}
exists, then \(f''(0)\) exists.
\begin{align*} \lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}&=\lim_{h \to 0} \frac{f'(h)-0}{h}=\lim_{h \to 0}\frac{f'(h)}{h}\\ \end{align*}Since \(f(h)\) behaves differently when \(h\) is greater than or less than zero, we look at the one-sided limits.
\begin{align*} \lim_{h \to 0^+}\frac{f'(h)}{h}&=\lim_{h \to 0^+}\frac{2h}{h}=2\\ \lim_{h \to 0^-}\frac{f'(h)}{h}&=\lim_{h \to 0^-}\frac{-2h}{h}=-2\\ \end{align*}Since the one-sided limits do not agree,
\begin{align*} \lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}&=DNE \end{align*}So, \(f''(0)\) does not exist. Now we have a complete picture of \(f''(x)\text{:}\)
\begin{equation*}
f''(x)=\left\{\begin{array}{ll}
-2&x \lt 0\\
DNE&x=0\\
2&x \gt 0.
\end{array}\right.
\end{equation*}
3 Applications of derivatives
3.1 Velocity and Acceleration
3.1.2 Exercises
3.1.2.1.
Answer.
False (but its velocity is zero)
3.1.2.2.
Answer.
It takes 10 seconds to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{,}\) and \(100\) seconds to accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\)
3.1.2.3.
Answer.
In general, false.
3.1.2.4.
Answer.
True
3.1.2.5.
Answer.
The pot is falling at 14 metres per second, just as it hits the ground.
3.1.2.6.
Answer.
(a) \(4.9x^2\) metres
(b) \(4.9x^2+x\) metres
3.1.2.7.
Answer.
\(8-4\sqrt{3}\approx 1\) sec
3.1.2.8.
Answer.
\(7.2\) sec
3.1.2.9.
Answer.
72 000 kph per hour
3.1.2.10.
Answer.
\(\dfrac{100}{7}\approx 14\) kph
3.1.2.11.
Answer.
about 1240 miles
3.1.2.12.
Answer.
\(49\) metres per second
3.1.2.13.
Answer.
About 416 metres
3.1.2.14.
Answer.
\(v_0=\sqrt{1960}\approx 44\) metres per second
3.1.2.15.
Answer.
\(\approx 74.2\) kph
3.1.2.16.
Answer.
Time elapsed: \(\dfrac{1}{4.9}+\dfrac{1}{9.8}\approx 0.3\) seconds
3.1.2.17.
Answer.
The acceleration is given by \(2^tv_0\log 2\text{,}\) where \(v_0\) is the velocity of the object at time \(t=0\text{.}\)
3.2 Related Rates
3.2.2 Exercises
3.2.2.1.
Answer.
ii and iv
3.2.2.2. (✳).
Answer.
\(-\dfrac{3}{2}\)
3.2.2.3. (✳).
Answer.
\(6\%\)
3.2.2.4. (✳).
3.2.2.5. (✳).
Answer.
\(-\dfrac{17}{5}\) units per second
3.2.2.6. (✳).
Answer.
\(\dfrac{4}{5}\) units per second
3.2.2.7. (✳).
Answer.
increasing at \(7\) mph
3.2.2.8. (✳).
Answer.
\(8\) cm per minute
3.2.2.9. (✳).
Answer.
\(-\dfrac{13}{6}\) metres per second
3.2.2.10.
Answer.
The height of the water is decreasing at \(\dfrac{3}{16}=0.1875\;\frac{\mathrm{cm}}{\mathrm{min}}\text{.}\)
3.2.2.11.
Answer.
\(\dfrac{1}{29200}\) metres per second (or about 1 centimetre every five minutes)
3.2.2.12.
Answer.
\(\left(\dfrac{2}{\left(\frac{1235}{72}\right)^2+4}\right)\left(\dfrac{6175}{3}\right)\approx
13.8\;\frac{\mathrm{rad}}{\mathrm{hour}} \approx 0.0038\;\frac{\mathrm{rad}}{\mathrm{sec}}\)
3.2.2.13. (✳).
Answer.
3.2.2.13.a \(\dfrac{24}{13}\approx 1.85\) km/min
3.2.2.13.b about \(.592\) radians/min
3.2.2.14.
Answer.
\(\dfrac{55\sqrt{21}\pi}{42}\approx 19\) centimetres per hour.
3.2.2.15. (✳).
Answer.
\(\ds\diff{A}{t}=-2\pi\;\dfrac{\mathrm{cm}^2}{\mathrm{s}}\)
3.2.2.16.
Answer.
\(288\pi\) cubic units per unit time
3.2.2.17.
Answer.
0 square centimetres per minute
3.2.2.18.
Answer.
\(-\dfrac{7\pi}{12}\approx -1.8\;\dfrac{\mathrm{cm}^3}{\mathrm{sec}^2}\)
3.2.2.19.
Answer.
The flow is decreasing at a rate of \(\dfrac{\sqrt{7}}{1000}\;\dfrac{\mathrm{m}^3}{\mathrm{sec}^2}\text{.}\)
3.2.2.20.
Answer.
\(\dfrac{-15}{49\pi}\approx -0.097\) cm per minute
3.2.2.21.
Answer.
- \(\ds\diff{D}{t}=\dfrac{1}{2\sqrt{2}}\) metres per hour
- The river is higher than 2 metres.
- The river’s flow has reversed direction. (This can happen near an ocean at high tide.)
3.2.2.22.
Answer.
(a) 2 units per second
(b) Its \(y\)-coordinate is decreasing at \(\dfrac{1}{2}\) unit per second. The point is moving at \(\dfrac{\sqrt{5}}{2}\) units per second.
3.2.2.23.
Answer.
(a) \(10\pi=\pi\left[3(a+b)-\sqrt{(a+3b)(3a+b)}\right]\) or equivalently, \(10=3(a+b)-\sqrt{(a+3b)(3a+b)}\)
(b) \(20\pi a b\)
(c) The water is spilling out at about 375.4 cubic centimetres per second. The exact amount is \(-\dfrac{200\pi}{9-\sqrt{35}}\left(1-2\left(\dfrac{3\sqrt{35}-11}{3\sqrt{35}-13}\right)\right)\;\dfrac{\mathrm{cm}^3}{\mathrm{sec}}\text{.}\)
3.2.2.24.
Answer.
\(B(10)=0\)
3.3 Exponential Growth and Decay — a First Look at Differential Equations
3.3.4 Exercises
Exercises for § 3.3.1
3.3.4.1.
Answer.
(a), (b)
3.3.4.2.
Answer.
(a), (d)
3.3.4.3.
Answer.
If \(C=0\text{,}\) then there was none to start out with, and \(Q(t)=0\) for all values of \(t\text{.}\)
If \(C \neq 0\text{,}\) then \(Q(t)\) will never be 0 (but as \(t\) gets bigger and bigger, \(Q(t)\) gets closer and closer to 0).
3.3.4.4. (✳).
Answer.
\(A=5\text{,}\) \(k = \dfrac{1}{7} \cdot \log\left(\pi/5\right)\)
3.3.4.5. (✳).
Answer.
\(y(t)=2e^{-3(t-1)}\text{,}\) or equivalently, \(y(t)=2e^3e^{-3t}\)
3.3.4.6.
Answer.
\(5\cdot 2^{-\tfrac{10000}{5730}}\approx 1.5\;\mu g\)
3.3.4.7.
Answer.
Radium-226 has a half life of about 1600 years.
3.3.4.8. (✳).
Answer.
\(\dfrac{\log 2}{\log 6}=\log_6(2)\) years, which is about 139 days
3.3.4.9.
Answer.
\(120\cdot\dfrac{\log 10}{\log 2}\) seconds, or about six and a half minutes.
3.3.4.10.
Answer.
About \(0.5\%\) of the sample decays in a day. The exact amount is \(\left[100\left(1-2^{-\tfrac{1}{138}}\right)\right] \%\text{.}\)
3.3.4.11.
Answer.
After ten years, the sample contains between 6.2 and 6.8 \(\mu\)g of Uranium-232.
Exercises for § 3.3.2
3.3.4.1.
Answer.
(a), (c), (d)
3.3.4.2.
Answer.
The temperature of the room is -10 degrees, and the room is colder than the object.
3.3.4.3.
Answer.
\(K\) is a negative number. It cannot be positive or zero.
3.3.4.4.
Answer.
If the object has a different initial temperature than its surroundings, then \(T(t)\) is never equal to \(A\text{.}\) (But as time goes on, it gets closer and closer.)
If the object starts out with the same temperature as its surrounding, then \(T(t)=A\) for all values of \(t\text{.}\)
3.3.4.5.
Answer.
\(\dfrac{-10\log(750)}{\log\left(\tfrac{2}{15}\right)}\approx 32.9\) seconds
3.3.4.6.
Answer.
\(10\dfrac{\log(10)}{\log(5)}\approx 14.3\) minutes
3.3.4.7. (✳).
Answer.
If Newton adds his cream just before drinking, the coffee ends up {cooler by \(0.85^\circ\) C}.
3.3.4.8. (✳).
Answer.
- \(\displaystyle \ds\diff{T}{t}=\frac{1}{5}\log\left(\frac{4}{5}\right)(T-30)\)
- \(\dfrac{5\log(2/5)}{\log(4/5)}\approx 20.53\) min
3.3.4.9.
Answer.
positive
Exercises for § 3.3.3
3.3.4.1.
Answer.
If \(P(0)=0\text{,}\) yes. If \(P(0) \neq 0\text{,}\) no: it does not take into account external constraints on population growth.
3.3.4.2.
Answer.
The Malthusian model predicts the herd will number 217 individuals in 2020.
3.3.4.3.
Answer.
\(\dfrac{\log(3)}{\log(2)}\approx 1.6\) hours
3.3.4.4.
Answer.
1912 or 1913
3.3.4.5.
Answer.
\(\dfrac{10^6}{5^4-1}\approx 1603\)
3.3.4.6.
Answer.
(a) At \(t=0\text{,}\) there are 100 units of the radioactive isotope in the sample. \(k\) is negative.
(b) At \(t=0\text{,}\) there are 100 individuals in the population. \(k\) is positive.
(c) The ambient temperature is 0 degrees. \(k\) is negative.
Further problems for § 3.3
3.3.4.1. (✳).
Answer.
\(f(2)=2e^{2\pi}\)
3.3.4.2.
Answer.
Solutions to the differential equation have the form
\begin{equation*}
T(t)=\left[T(0)+\frac{9}{7}\right]e^{7t}-\frac{9}{7}
\end{equation*}
for some constant \(T(0)\text{.}\)
3.3.4.3. (✳).
Answer.
\(\dfrac{8\log(0.4)}{\log(0.8)}\approx 32.85\) days
3.3.4.4.
Answer.
25\(^\circ\) C
3.3.4.5. (✳).
Answer.
- \(A(t)=90,\!000\cdot e^{0.05t}-40,\!000\text{.}\) When the graduate is 65, they will have $625,015.05 in the account.
- $49,437.96
3.3.4.6. (✳).
3.3.4.7. (✳).
Answer.
\(\dfrac{9\log 2}{\log 3}\approx5.68\) hr
3.3.4.8. (✳).
Answer.
(a) \(v(t)=\left[v_0+\frac{g}{k}\right]e^{-kt}-\frac{g}{k}\)
(b) \(\lim\limits_{t\rightarrow\infty} v(t)=-\dfrac{g}{k}\)
3.4 Approximating Functions Near a Specified Point — Taylor Polynomials
3.4.11 Exercises
Exercises for § 3.4.1
3.4.11.1.
Answer.
Since \(f(0)\) is closer to \(g(0)\) than it is to \(h(0)\text{,}\) you would probably want to estimate \(f(0) \approx g(0)=1+2\sin (1)\) if you had the means to efficiently figure out what \(\sin(1)\) is, and if you were concerned with accuracy. If you had a calculator, you could use this estimation. Also, later in this chapter we will learn methods of approximating \(\sin (1)\) that do not require a calculator, but they do require time.
Without a calculator, or without a lot of time, using \(f(0)\approx h(0)=0.7\) probably makes the most sense. It isn’t as accurate as \(f(0) \approx g(0)\text{,}\) but you get an estimate very quickly, without worrying about figuring out what \(\sin(1)\) is.
3.4.11.2.
Answer.
\(\log(0.93)\approx \log(1)=0\)
3.4.11.3.
Answer.
\(\arcsin(0.1) \approx 0\)
3.4.11.4.
Answer.
\(\sqrt{3}\tan(1) \approx 3\)
3.4.11.5.
Answer.
\(10.1^3 \approx 10^3=1000\)
Exercises for § 3.4.2
3.4.11.1.
Answer.
(a) \(f(5)=6\)
(b) \(f'(5)=3\)
(c) not enough information to know
3.4.11.2.
Answer.
The linear approximation is shown in red.
3.4.11.3.
Answer.
\(f(x)=2x+5\)
3.4.11.4.
Answer.
\(\log(0.93) \approx -0.07\)
3.4.11.5.
Answer.
\(\sqrt{5} \approx \dfrac{9}{4}\)
3.4.11.6.
Answer.
\(\sqrt[5]{30}\approx \dfrac{79}{40}\)
3.4.11.7.
Answer.
\(10.1^3 \approx 1030\text{,}\) \(10.1^3 = 1030.301\)
3.4.11.8.
Answer.
There are many possible answers. One is \(f(x)=\sin x\text{,}\) \(a=0\text{,}\) and \(b=\pi\text{.}\)
3.4.11.9.
Answer.
\(a=\sqrt{3}\)
Exercises for § 3.4.3
3.4.11.1.
Answer.
\(f(3)=9\text{,}\) \(f'(3)=0\text{,}\) \(f''(3)=-2\text{;}\) there is not enough information to know \(f'''(3)\text{.}\)
3.4.11.2.
Answer.
\(f(x) \approx 2x+5\)
3.4.11.3.
Answer.
\(\log(0.93) \approx -0.07245\)
3.4.11.4.
Answer.
\(\cos\left(\dfrac{1}{15}\right )\approx \dfrac{449}{450}\)
3.4.11.5.
Answer.
\(e^{2x}\approx 1+2x+2x^2\)
3.4.11.6.
Answer.
One approximation: \(5^{\tfrac{4}{3}}\approx\dfrac{275}{32}\)
3.4.11.7.
Answer.
3.4.11.8.
Answer.
For each of these, there are many solutions. We provide some below.
- \(\displaystyle 1+2+3+4+5 = \ds\sum_{n=1}^5n\)
- \(\displaystyle 2+4+6+8=\ds\sum_{n=1}^42n\)
- \(\displaystyle 3+5+7+9+11=\ds\sum_{n=1}^{5}(2n+1)\)
- \(\displaystyle 9+16+25+36+49=\ds\sum_{n=3}^7 n^2\)
- \(\displaystyle 9+4+16+5+25+6+36+7+49+8=\ds\sum_{n=3}^7 (n^2+n+1)\)
- \(\displaystyle 8+15+24+35+48= \ds\sum_{n=3}^7 (n^2-1)\)
- \(\displaystyle 3-6+9-12+15-18=\ds\sum_{n=1}^6 (-1)^{n+1}3n\)
3.4.11.9.
Answer.
\(f(1) \approx 2\text{,}\) \(f(1)=\pi\)
3.4.11.10.
Answer.
\(e\approx 2.5\)
3.4.11.11.
Answer.
- [3.4.11.11.b=3.4.11.11.g], and
Exercises for § 3.4.4
3.4.11.1.
Answer.
\(f''(1)=-4\)
3.4.11.2.
Answer.
\(f^{(10)}(5)=10!\)
3.4.11.3.
Answer.
\(T_3(x)=-x^3+x^2-x+1\)
3.4.11.4.
Answer.
\(T_3(x)=-7+7(x-1)+9(x-1)^2+5(x-1)^3\text{,}\) or equivalently, \(T_3(x)=5x^3-6x^2+4x-10\)
3.4.11.5.
Answer.
\(f^{(10)}(5)=\dfrac{11\cdot 10!}{6}\)
3.4.11.6.
Answer.
\(a=\sqrt{e}\)
Exercises for § 3.4.5
3.4.11.1.
Answer.
\begin{align*}
T_{16}(x)&=1\textcolor{blue}{+}x
\textcolor{red}{-}\frac{1}{2}x^2
\textcolor{red}{-}\frac{1}{3!}x^3
\textcolor{blue}{+}\frac{1}{4!}x^4
\textcolor{blue}{+}\frac{1}{5!}x^5
\textcolor{red}{-}\frac{1}{6!}x^6
\textcolor{red}{-}\frac{1}{7!}x^7
\textcolor{blue}{+}\frac{1}{8!}x^8
\textcolor{blue}{+}\frac{1}{9!}x^9\\
\amp\qquad\textcolor{red}{-}\frac{1}{10!}x^{10}
\textcolor{red}{-}\frac{1}{11!}x^{11}
\textcolor{blue}{+}\frac{1}{12!}x^{12}
\textcolor{blue}{+}\frac{1}{13!}x^{13}
\textcolor{red}{-}\frac{1}{14!}x^{14}
\textcolor{red}{-}\frac{1}{15!}x^{15}\\
&\qquad\textcolor{blue}{+}\frac{1}{16!}x^{16}
\end{align*}
3.4.11.2.
Answer.
\(T_{100}(t)=127.5+48(t-5)+4.9(t-5)^2=4.9t^2-t+10\)
3.4.11.3.
Answer.
\(T_n(x)=\ds\sum_{k=0}^n \frac{2(\log 2)^k}{k!}(x-1)^k\)
3.4.11.4.
Answer.
\begin{align*}
T_6(x)\amp=7+5(x-1)+\frac{7}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{12}(x-1)^4\\
\amp\qquad+\frac{1}{30}(x-1)^5-\frac{1}{60}(x-1)^6
\end{align*}
3.4.11.5.
Answer.
\(T_n(x)=\ds\sum_{k=0}^n x^k\)
3.4.11.6.
Answer.
\(T_3(x)=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3\)
3.4.11.7.
Answer.
\(\pi=6\arctan\left(\dfrac{1}{\sqrt3}\right)\approx\dfrac{82}{45}\sqrt{3}\approx3.156\)
3.4.11.8.
Answer.
\(T_{100}(x)=-1+\ds\sum_{k=2}^{100}\frac{(-1)^k}{k(k-1)}(x-1)^k\)
3.4.11.9.
Answer.
\begin{equation*}
T_{2n}(x)=\sum_{\ell=0}^{n}\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell}
+\sum_{\ell=0}^{n-1}\frac{(-1)^\ell}{(2\ell+1)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell+1}
\end{equation*}
3.4.11.10.
Answer.
\begin{equation*}
1+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{157!}\approx e-1
\end{equation*}
3.4.11.11.
Answer.
We estimate that the sum is close to \(-\dfrac{1}{\sqrt{2}}\text{.}\)
Exercises for § 3.4.6
3.4.11.1.
Answer.
3.4.11.2.
Answer.
Let \(f(x)\) be the number of problems finished after \(x\) minutes of work. The question tells us that \(5\Delta y \approx \Delta x\text{.}\) So, if \(\Delta x = 15\text{,}\) \(\Delta y \approx 3\text{.}\) That is, in 15 minutes more, you will finish about 3 more problems.
3.4.11.3.
Answer.
(a) \(\Delta y \approx \dfrac{1}{260}\approx 0.003846\)
(b) \(\Delta y \approx \dfrac{51}{13520}\approx 0.003772\)
3.4.11.4.
Answer.
(a) \(\Delta y \approx 1.1\) metres per second
(b) The increase from the first to the second jump is bigger.
Exercises for § 3.4.7
3.4.11.1.
Answer.
False.
3.4.11.2.
Answer.
Absolute error: 0.17; percentage error: 2.92%
3.4.11.3.
Answer.
The linear approximates estimates the error in \(f(x)\) to be about 60, while the quadratic approximates estimates the error in \(f(x)\) to be about 63.
3.4.11.4.
Answer.
1%
3.4.11.5.
Answer.
(a) \(\dfrac{9}{2}\theta\)
(b) \(\theta = 2\arcsin\left(\dfrac{d}{6}\right)\)
(c) \(\dfrac{9}{\sqrt{36-0.68^2}}\cdot0.02\approx 0.03\)
3.4.11.6.
Answer.
We estimate that the volume decreased by about \(0.00245\) cubic metres, or about 2450 cubic centimetres.
3.4.11.7.
Answer.
Correct to within about 10.4 years (or about 53%)
Exercises for § 3.4.8
3.4.11.1.
Answer.
(a) False
(b) True
(c) True
(d) True
3.4.11.2.
Answer.
Equation 3.4.33 gives us the bound \(|f(2)-T_3(2)| \lt 6\text{.}\) A calculator tells us actually \(|f(2)-T_3(2)|\approx 1.056\text{.}\)
3.4.11.3.
Answer.
\(|f(37)-T(37)|=0\)
3.4.11.4.
Answer.
You do, you clever goose!
3.4.11.5.
Answer.
\(|f(11.5)-T_5(11.5)| \lt \dfrac{9}{7\cdot 2^6} \lt 0.02\)
3.4.11.6.
Answer.
\(\left|f(0.1)-T_2(0.1)\right| \lt \dfrac{1}{1125}\)
3.4.11.7.
Answer.
\(\left|f\left(-\dfrac{1}{4}\right)-T_5\left(-\dfrac{1}{4}\right)\right| \lt \dfrac{1}{6\cdot 4^6} \lt 0.00004\)
3.4.11.8.
Answer.
Your answer may vary. One reasonable answer is \(\left|f(30)-T_3(30)\right| \lt \dfrac{14}{5^7\cdot9\cdot15} \lt 0.000002\text{.}\)
Another reasonable answer is \(\left|f(30)-T_3(30)\right| \lt \dfrac{14}{5^7\cdot9} \lt 0.00002\text{.}\)
3.4.11.9.
Answer.
Equation 3.4.33 gives the bound \(|f(0.01)-T_n(0.01)| \leq 100^2\left(\frac{100}{\pi}-1\right)^2\text{.}\)
A more reasonable bound on the error is that it is less than 5.
3.4.11.10.
Answer.
Using Equation 3.4.33,
\begin{align*}
\left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right|& \lt \frac{1}{10}.\\
\end{align*}
The actual error is
\begin{align*} \left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right|&= \frac{\pi}{6}-\frac{1}{2} \end{align*}which is about 0.02.
3.4.11.11.
Answer.
Any \(n\) greater than or equal to 3.
3.4.11.12.
Answer.
\(\sqrt[7]{2200}\approx3+\dfrac{13}{7\cdot 3^6}
\approx 3.00255\)
3.4.11.13.
Answer.
If we’re going to use Equation 3.4.33, then we’ll probably be taking a Taylor polynomial. Using Example 3.4.16, the 6th order Maclaurin polynomial for \(\sin x\) is
\begin{equation*}
T_6(x)=T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}
\end{equation*}
so let’s play with this a bit. Equation 3.4.33 tells us that the error will depend on the seventh derivative of \(f(x)\text{,}\) which is \(-\cos x\text{:}\)
\begin{align*}
f(1)-T_6(1)&=f^{(7)}(c)\frac{1^7}{7!}\\
\sin(1)-\left(1-\frac{1}{3!}+\frac{1}{5!}\right)&=\frac{-\cos c}{7!}\\
\sin(1)-\frac{101}{5!}&=\frac{-\cos c}{7!}\\
\sin(1)&=\frac{4242-\cos c}{7!}\\
\end{align*}
for some \(c\) between 0 and 1. Since \(-1 \leq \cos c \leq 1\text{,}\)
\begin{align*} \frac{4242-1}{7!}\leq \sin(1)&\leq \frac{4242+1}{7!}\\ \frac{4241}{7!}\leq \sin(1)&\leq \frac{4243}{7!}\\ \frac{4241}{5040}\leq \sin(1)&\leq \frac{4243}{5040} \end{align*}Remark: there are lots of ways to play with this idea to get better estimates. One way is to take a higher order Maclaurin polynomial. Another is to note that, since \(0 \lt c \lt 1 \lt \dfrac{\pi}{3}\text{,}\) then \(\dfrac{1}{2} \lt \cos c \lt 1\text{,}\) so
\begin{gather*}
\dfrac{4242-1}{7!} \lt \sin(1) \lt \dfrac{4242-\frac{1}{2}}{7!}\\
\dfrac{4241}{5040} \lt \sin(1) \lt \dfrac{8483}{10080} \lt \frac{4243}{5040}
\end{gather*}
If you got tighter bounds than asked for in the problem, congratulations!
3.4.11.14.
Answer.
(a) \(T_4(x)=\sum_{n=0}^4\frac{x^n}{n!}\)
(b) \(T_4(1)=\frac{65}{24}\)
(c) See the solution.
Further problems for § 3.4
3.4.11.1. (✳).
Answer.
\(f'(0)=0\) and \(f''(0)=6\text{.}\)
3.4.11.2. (✳).
Answer.
\(4\)
3.4.11.3. (✳).
Answer.
\(h'(2)=\dfrac{1}{2}\text{,}\) \(h''(2)=0\)
3.4.11.4. (✳).
Answer.
(a) \(1.92\)
(b) \(1.918\)
3.4.11.5. (✳).
Answer.
\(10^{1/3}\approx\dfrac{13}{6}\text{;}\) this approximation is too big.
3.4.11.6. (✳).
Answer.
\(\sqrt{2}\approx \dfrac{3}{2}\)
3.4.11.7. (✳).
Answer.
\(\sqrt[3]{26}\approx \dfrac{80}{27}\)
3.4.11.8. (✳).
Answer.
\((10.1)^5 \approx 105,000\)
3.4.11.9. (✳).
Answer.
\(\sin\left(\dfrac{101\pi}{100}\right) \approx -\dfrac{\pi}{100}\)
3.4.11.10. (✳).
Answer.
\(\arctan(1.1)\approx\left(\dfrac{\pi}{4}+\dfrac{1}{20}\right)\)
3.4.11.11. (✳).
Answer.
\(\dfrac{8012}{1000}\)
3.4.11.12. (✳).
Answer.
\((8.06)^{2/3}\approx\dfrac{402}{100}=\dfrac{201}{50}\)
3.4.11.13. (✳).
Answer.
\(1 + x +2x^2+\dfrac{14}{3}x^3\)
3.4.11.14. (✳).
Answer.
- By Equation 3.4.33, the absolute value of the error is\begin{equation*} \left|\frac{f'''(c)}{3!}\cdot (2-1)^3\right| = \left|\frac{c}{6(22-c^2)}\right| \end{equation*}for some \(c \in (1,2)\text{.}\)
- When \(1\leq c\leq2\text{,}\) we know that \(18 \leq 22-c^2 \leq 21\text{,}\) and that numerator and denominator are non-negative, so\begin{align*} \left|\frac{c}{6(22-c^2)}\right| &=\frac{c}{6(22-c^2)} \leq \frac{2}{6(22-c^2)} \leq \frac{2}{6\cdot 18}\\ & = \frac{1}{54} \leq \frac{1}{50} \end{align*}as required.
- Alternatively, notice that \(c\) is an increasing function of \(c\text{,}\) while \(22-c^2\) is a decreasing function of \(c\text{.}\) Hence the fraction is an increasing function of \(c\) and takes its largest value at \(c=2\text{.}\) Hence\begin{align*} \left|\frac{c}{6(22-c^2)}\right| & \leq \frac{2}{6\times 18} = \frac{1}{54} \leq \frac{1}{50}. \end{align*}
3.4.11.15. (✳).
Answer.
- By Equation 3.4.33, there is \(c\in(0,0.5)\) such that the error is\begin{align*} R_4 &= \frac{f^{(4)}(c)}{4!} (0.5-0)^4\\ &= \frac{1}{24\cdot 16} \cdot \frac{\cos(c^2)}{3-c} \end{align*}
- For any \(c\) we have \(|\cos(c^2)| \leq 1\text{,}\) and for \(c \lt 0.5\) we have \(3-c \gt 2.5\text{,}\) so that\begin{equation*} \left|\frac{\cos(c^2)}{3-c}\right| \leq \frac{1}{2.5}\,. \end{equation*}
- We conclude that\begin{equation*} \left| R_4 \right| \leq \frac{1}{2.5\cdot 24\cdot 16} = \frac{1}{60\cdot 16} \lt \frac{1}{60\cdot 10}=\frac{1}{600} \lt \frac{1}{500} \end{equation*}
3.4.11.16. (✳).
Answer.
- By Equation 3.4.33, there is \(c\in(0,1)\) such that the error is\begin{align*} \left|\frac{f'''(c)}{3!}\cdot (1-0)^3\right| &= \left|\frac{e^{-c}}{6(8+c^2)}\right|. \end{align*}
- When \(0 \lt c \lt 1\text{,}\) we know that \(1 \gt e^{-c} \gt e^{-1}\) and \(8 \leq 8+c^2 \lt 9\text{,}\) so\begin{align*} \left|\frac{e^{-c}}{6(8+c^2)}\right| &=\frac{e^{-c}}{6(8+c^2)}\\ & \lt \frac{1}{6 |8+c^2|}\\ & \lt \frac{1}{6\times 8} = \frac{1}{48} \lt \frac{1}{40} \end{align*}as required.
3.4.11.17. (✳).
Answer.
3.4.11.18.
Answer.
\(T_3(x)=5x^2-9\)
3.4.11.19. (✳).
Answer.
3.4.11.20. (✳).
Answer.
3.4.11.21. (✳).
Answer.
\(\log 10.3 \approx 2.33259\) The error is between \(-0.00045\) and \(-0.00042\text{.}\)
3.4.11.22. (✳).
Answer.
(a) \(L(x)=e+ex\)
(b) \(Q(x)=e+ex+ex^2\)
(c) Since \(ex^2 \gt 0\) for all \(x \gt 0\text{,}\) \(L(x) \lt Q(x)\) for all \(x \gt 0\text{.}\)
From the error formula, we know that
\begin{align*}
f(x)&=f(0)+f'(0)x+\half f''(0)x^2
+\dfrac{1}{3!}f'''(c)x^{3}\\
&=Q(x)+\frac{1}{6}\left(e^c+3e^{2c}+e^{3c}\right)e^{e^c}x^3
\end{align*}
for some \(c\) between \(0\) and \(x\text{.}\) Since \(\frac{1}{6}\left(e^c+3e^{2c}+e^{3c}\right)e^{e^c}\) is positive for any \(c\text{,}\) for all \(x \gt 0\text{,}\) \(\frac{1}{6}\left(e^c+3e^{2c}+e^{3c}\right)e^{e^c}x^3 \gt 0\text{,}\) so \(Q(x) \lt f(x)\text{.}\)
(d) \(1.105 \lt e^{0.1} \lt 1.115\)
3.5 Optimisation
3.5.4 Exercises
Exercises for § 3.5.1
3.5.4.1.
Answer.
There is a critical point at \(x=0\text{.}\) The \(x\)-value of the red dot is a singular point, and a local maximum occurs there.
3.5.4.2.
Answer.
The \(x\)-coordinate corresponding to the blue dot (let’s call it \(a\)) is a critical point, and \(f(x)\) has a local and global minimum at \(x=a\text{.}\) The \(x\)-coordinate corresponding to the discontinuity (let’s call it \(b\)) is a singular point, but there is not a global or local extremum at \(x=b\text{.}\)
3.5.4.3.
Answer.
One possible answer is shown below.
3.5.4.4.
Answer.
The critical points are \(x=3\) and \(x=-1\text{.}\) These two points are the only places where local extrema might exist. There are no singular points.
3.5.4.5.
Answer.
3.5.4.6.
Answer.
There are many possible answers. Every answer must have \(x=2\) as a singular point strictly inside the domain of \(f(x)\text{.}\) Two possibilities are shown below.
3.5.4.7.
Answer.
\(x=-7\text{,}\) \(x=-1\text{,}\) and \(x=5\)
3.5.4.8.
Answer.
Every real number \(c\) is a critical point of \(f(x)\text{,}\) and \(f(x)\) has a local and global maximum and minimum at \(x=c\text{.}\) There are no singular points.
Exercises for § 3.5.2
3.5.4.1.
Answer.
Two examples are given below, but many are possible.
3.5.4.2.
Answer.
Two examples are given below, but many are possible.
3.5.4.3.
Answer.
One possible answer:
3.5.4.4.
Answer.
The global maximum is 45 at \(x=5\) and the global minimum is \(-19\) at \(x=-3\text{.}\)
3.5.4.5.
Answer.
The global maximum over the interval is \(61\) at \(x=-3\text{,}\) and the global minimum is \(7\) at \(x=0\text{.}\)
Exercises for § 3.5.3
3.5.4.1. (✳).
Answer.
The global maximum is \(f(-1) = 6\text{,}\) the global minimum is \(f(-2) = -20\text{.}\)
3.5.4.2. (✳).
Answer.
Global maximum is \(f(2) = 12\text{,}\) global minimum is \(f(1) = -14\text{.}\)
3.5.4.3. (✳).
Answer.
Global maximum is \(f(4) = 30\text{,}\) global minimum is \(f(2) = -10\text{.}\)
3.5.4.4. (✳).
Answer.
Local max at \((-2,20)\text{,}\) local min at \((2,-12)\text{.}\)
3.5.4.5. (✳).
Answer.
\((-2,33)\) max, and \((2,-31)\) min
3.5.4.6. (✳).
Answer.
\(Q\) should be \(4\sqrt{3}\) kilometres from \(A\)
3.5.4.7. (✳).
Answer.
\(10\times 30 \times 15\)
3.5.4.8. (✳).
Answer.
\(2\times 2\times 6\)
3.5.4.9. (✳).
Answer.
\(X=Y=\sqrt{2}\)
3.5.4.10. (✳).
Answer.
The largest possible perimeter is \(2\sqrt{5}R\) and the smallest possible perimeter is \(2R\text{.}\)
3.5.4.11. (✳).
Answer.
\(\dfrac{A^{3/2}}{3\sqrt{6\pi}}\)
3.5.4.12. (✳).
Answer.
\(\dfrac{P^2}{2(\pi+4)}\)
3.5.4.13. (✳).
Answer.
(a) \(x=\sqrt{\dfrac{A}{3p}}\text{,}\) \(y=\sqrt{\dfrac{Ap}{3}}\text{,}\) and \(z=\dfrac{\sqrt{Ap}}{\sqrt{3}(1+p)}\)
(b) \(p=1\)
(The dimensions of the resulting baking pan are \(x=y=\sqrt{ \dfrac{A}{3} }\) and \(z=\dfrac{1}{2}\sqrt{ \dfrac{A}{3} }\text{.}\))
3.5.4.14. (✳).
Answer.
3.5.4.15. (✳).
Answer.
Maximum area: do not cut, make a circle and no square.
Minimum area: make a square out of a piece that is \(\dfrac{4}{4+\pi}\) of the total length of the wire.
3.6 Sketching Graphs
3.6.7 Exercises
Exercises for § 3.6.1
3.6.7.1.
Answer.
In general, false.
3.6.7.2.
Answer.
\(f(x)=A(x)\)
\(g(x)=C(x)\)
\(h(x)=B(x)\)
\(k(x)=D(x)\)
3.6.7.3.
Answer.
(a) \(p=e^2\)
(b) \(b=-e^2\text{,}\) \(1-e^2\)
3.6.7.4.
Answer.
vertical asymptote at \(x=3\text{;}\) horizontal asymptotes \(\ds\lim_{x \to \pm \infty}f(x)=\dfrac{2}{3}\)
3.6.7.5.
Answer.
horizontal asymptote \(y=0\) as \(x \to -\infty\text{;}\) no other asymptotes
Exercises for § 3.6.2
3.6.7.1.
Answer.
\begin{align*}
\amp\textcolor{green}{A'(x)=l(x)} \amp
\amp\textcolor{blue}{B'(x)=p(x)} \amp
\amp\textcolor{red}{C'(x)=n(x)}\\
\amp\textcolor{orange}{D'(x)=o(x)}\amp\
\amp\textcolor{purple}{E'(x)=m(x)}
\end{align*}
3.6.7.2. (✳).
Answer.
\((-2,\infty)\)
3.6.7.3. (✳).
Answer.
\((1,4)\)
3.6.7.4. (✳).
Answer.
\((-\infty,1)\)
Exercises for § 3.6.3
3.6.7.1.
Answer.
3.6.7.2.
Answer.
3.6.7.3.
Answer.
In general, false.
3.6.7.4. (✳).
Answer.
\(x=1,\ y=11\)
3.6.7.5. (✳).
Answer.
Let
\begin{equation*}
g(x)=f''(x)=x^3+5x-20.
\end{equation*}
Then \(g'(x)=3x^2+5\text{,}\) which is always positive. That means \(g(x)\) is strictly increasing for all \(x\text{.}\) So, \(g(x)\) can change signs once, from negative to positive, but it can never change back to negative. An inflection point of \(f(x)\) occurs when \(g(x)\) changes signs. So, \(f(x)\) has at most one inflection point.
Since \(g(x)\) is continuous, we can apply the Intermediate Value Theorem to it. Notice \(g(3) \gt 0\) while \(g(0) \lt 0\text{.}\) By the IVT, \(g(x)=0\) for at least one \(x \in (0,3)\text{.}\) Since \(g(x)\) is strictly increasing, at the point where \(g(x)=0\text{,}\) \(g(x)\) changes from negative to positive. So, the concavity of \(f(x)\) changes. Therefore, \(f(x)\) has at least one inflection point.
Now that we’ve shown that \(f(x)\) has at most one inflection point, and at least one inflection point, we conclude it has exactly one inflection point.
3.6.7.6. (✳).
Answer.
3.6.7.6.a Let
\begin{equation*}
g(x)=f'(x)
\end{equation*}
Then \(f''(x)\) is the derivative of \(g(x)\text{.}\) Since \(f''(x) \gt 0\) for all \(x\text{,}\) \(g(x)=f'(x)\) is strictly increasing for all \(x\text{.}\) In other words, if \(y \gt x\) then \(g(y) \gt g(x)\text{.}\)
Suppose \(g(x)=0\text{.}\) Then for every \(y\) that is larger than \(x\text{,}\) \(g(y) \gt g(x)\text{,}\) so \(g(y) \neq 0\text{.}\) Similarly, for every \(y\) that is smaller than \(x\text{,}\) \(g(y) \lt g(x)\text{,}\) so \(g(y) \neq 0\text{.}\) Therefore, \(g(x)\) can only be zero for at most one value of \(x\text{.}\) Since \(g(x)=f'(x)\text{,}\) that means \(f(x)\) can have at most one critical point.
Suppose \(f'(c)=0\text{.}\) Since \(f'(x)\) is a strictly increasing function, \(f'(x) \lt 0\) for all \(x \lt c\) and \(f'(x) \gt 0\) for all \(x \gt c\text{.}\)
Then \(f(x)\) is decreasing for \(x \lt c\) and increasing for \(x \gt c\text{.}\) So \(f(x) \gt f(c)\) for all \(x\neq c\text{.}\)
Since \(f(x) \gt f(c)\) for all \(x\ne c\text{,}\) so \(c\) is an absolute minimum for \(f(x)\text{.}\)
3.6.7.6.b We know that the maximum over an interval occurs at an endpoint, a critical point, or a singular point.
- Since \(f'(x)\) exists everywhere, there are no singular points.
- If the maximum were achieved at a critical point, that critical point would have to provide both the absolute maximum and the absolute minimum (by part (a)). So, the function would have to be a constant and consequently could not have a nonzero second derivative. So the maximum is not at a critical point.
That leaves only the endpoints of the interval.
3.6.7.7.
Answer.
If \(x=3\) is an inflection point, then the concavity of \(f(x)\) changes at \(x=3\text{.}\) That is, there is some interval strictly containing 3, with endpoints \(a\) and \(b\text{,}\) such that
- \(f''(a) \lt 0\) and \(f''(x) \lt 0\) for every \(x\) between \(a\) and 3, and
- \(f''(b) \gt 0\) and \(f''(x) \gt 0\) for every \(x\) between \(b\) and 3.
Since \(f''(a) \lt 0\) and \(f''(b) \gt 0\text{,}\) and since \(f''(x)\) is continuous, the Intermediate Value Theorem tells us that there exists some \(x\) strictly between \(a\) and \(b\) with \(f''(x)=0\text{.}\) So, we know \(f''(x)=0\) somewhere between \(a\) and \(b\text{.}\) The question is, where exactly could that be?
- \(f''(x) \lt 0\) (and hence \(f''(x) \neq 0\)) for all \(x\) between \(a\) and 3
- \(f''(x) \gt 0\) (and hence \(f''(x) \neq 0\)) for all \(x\) between \(b\) and 3
- So, any number between \(a\) and \(b\) that is not 3 has \(f''(x)\neq 0\text{.}\)
So, \(x=3\) is the only possible place between \(a\) and \(b\) where \(f''(x)\) could be zero. Therefore, \(f''(3)=0\text{.}\)
Exercises for § 3.6.4
3.6.7.1.
Answer.
even
3.6.7.2.
Answer.
odd, periodic
3.6.7.3.
Answer.
3.6.7.4.
Answer.
3.6.7.5.
Answer.
A function is even if \(f(-x)=f(x)\text{.}\)
\begin{align*}
f(-x)&=\frac{(-x)^4-(-x)^6}{e^{(-x)^2}}\\
&=\frac{x^4-x^6}{e^{x^2}}\\
&=f(x)
\end{align*}
So, \(f(x)\) is even.
3.6.7.6.
Answer.
For any real number \(x\text{,}\) we will show that \(f(x)=f(x+4\pi)\text{.}\)
\begin{align*}
f(x+4\pi)&=\sin(x+4\pi)+\cos\left(\frac{x+4\pi}{2}\right)\\
&=\sin(x+4\pi)+\cos\left(\frac{x}{2}+2\pi\right)\\
&=\sin(x)+\cos\left(\frac{x}{2}\right)\\
&=f(x)
\end{align*}
So, \(f(x)\) is periodic.
3.6.7.7.
Answer.
even
3.6.7.8.
Answer.
none
3.6.7.9.
Answer.
1
3.6.7.10.
Answer.
\(\pi\)
Exercises for § 3.6.6
3.6.7.1. (✳).
Answer.
3.6.7.2. (✳).
Answer.
The open dot is the inflection point, and the closed dot is the local and global maximum.
3.6.7.3. (✳).
Answer.
The open dot marks the inflection point.
3.6.7.4. (✳).
Answer.
3.6.7.5. (✳).
Answer.
3.6.7.5.a One branch of the function, the exponential function \(e^x\text{,}\) is continuous everywhere. So \(f(x)\) is continuous for \(x \lt 0\text{.}\) When \(x \geq 0\text{,}\) \(f(x)=\dfrac{x^2+3}{3(x+1)}\text{,}\) which is continuous whenever \(x \neq -1\) (so it’s continuous for all \(x \gt 0\)). So, \(f(x)\) is continuous for \(x \gt 0\text{.}\) To see that \(f(x)\) is continuous at \(x=0\text{,}\) we see:
\begin{align*}
\lim_{x\rightarrow0-}f(x)=\lim_{x\rightarrow0-}e^x&=1\\
\lim_{x\rightarrow0+}f(x)=\lim_{x\rightarrow0+}\frac{x^2+3}{3(x+1)}&=1\\
\mbox{So, } \lim_{x \rightarrow 0}f(x)&=1=f(0)
\end{align*}
Hence \(f(x)\) is continuous at \(x=0\text{,}\) so \(f(x)\) is continuous everywhere.
- i. \(f(x)\) is increasing for \(x \lt 0\) and \(x \gt 1\text{,}\) decreasing for \(0 \lt x \lt 1\text{,}\) has a local max at \((0,1)\text{,}\) and has a local min at \(\left(1,\frac{2}{3}\right)\text{.}\)
- ii. \(f(x)\) is concave upwards for all \(x\ne 0\text{.}\)
- iii. The \(x\)--axis is a horizontal asymptote as \(x\rightarrow-\infty\text{.}\)
3.6.7.6. (✳).
Answer.
3.6.7.7. (✳).
Answer.
- Increasing: \((-1,1)\text{,}\) decreasing: \((-\infty,-1)\cup (1,\infty)\)
- concave up: \((-\sqrt{3},0) \cup (\sqrt{3},\infty)\text{,}\) concave down: \((-\infty,-\sqrt{3}) \cup (0 ,\sqrt{3})\)
- inflection points: \(x=\pm\sqrt{3}, 0\)
3.6.7.7.b The local and global minimum of \(f(x)\) is at \((-1,\frac{-1}{\sqrt{e}})\text{,}\) and the local and global maximum of \(f(x)\) is at \((1,\frac{1}{\sqrt{e}})\text{.}\)
3.6.7.7.c In the graph below, open dots are inflection points, and solid dots are extrema.
3.6.7.8.
Answer.
Local maxima occur at \(x=\frac{2\pi}{3}+2\pi n\) for all integers \(n\text{,}\) and local minima occur at \(x=-\frac{2\pi}{3}+2\pi n\) for all integers \(n\text{.}\) Inflection points occur at every integer multiple of \(\pi\text{.}\)
3.6.7.9. (✳).
Answer.
Below is the graph \(y=f(x)\) over the interval \([-\pi,\pi]\text{.}\) The sketch of the curve over a larger domain is simply a repetition of this figure.
On the interval \([0,\pi]\text{,}\) the maximum value of \(f(x)\) is \(6\) and the minimum value is \(-2\text{.}\)
Let \(a=\arcsin \left(\dfrac{-1+\sqrt{33}}{8}\right)\approx 0.635\approx0.2\pi\) and \(b=\arcsin \left(\dfrac{-1-\sqrt{33}}{8}\right)\approx -1.003\approx -0.3\pi\). The points \(-\pi-b\), \(b\), \(a\), and \(\pi-a\) are inflection points.
3.6.7.10.
Answer.
The closed dot is the local minimum, and the open dots are inflection points at \(x=-1\) and \(x=-2\pm\sqrt{1.5}\text{.}\) The graph has horizontal asymptotes \(y=0\) as \(x\) goes to \(\pm \infty\text{.}\)
3.6.7.11. (✳).
Answer.
- 3.6.7.11.a decreasing for \(x \lt 0\) and \(x \gt 2\text{,}\) increasing for \(0 \lt x \lt 2\text{,}\) minimum at \((0,0)\text{,}\) maximum at \((2,2)\text{.}\)
- 3.6.7.11.b concave up for \(x \lt 2-\sqrt{2}\) and \(x \gt 2+\sqrt{2}\text{,}\) concave down for \(2-\sqrt 2 \lt x \lt 2+\sqrt 2\text{,}\) inflection points at \(x = 2\pm \sqrt{2}\text{.}\)
- 3.6.7.11.c\(\infty\)
Open dots indicate inflection points, and closed dots indicate local extrema.
3.6.7.12. (✳).
Answer.
There are no inflection points or extrema, except the endpoint \((0,1)\text{.}\)
There are no inflection points or extrema, except the endpoint \((1,0)\text{.}\)
3.6.7.12.d \(g'(\half)=-2\)
3.6.7.13. (✳).
Answer.
(a)
Local maximum at \(x=-\frac{1}{\sqrt[4]{5}}\text{;}\) local minimum at \(x=\frac{1}{\sqrt[4]{5}}\text{;}\) inflection point at the origin; concave down for \(x\lt 0\) ; concave up for \(x\gt 0\text{.}\)
(b) The number of distinct real roots of \(x^5-x+k\) is:
- 1 when \(|k| \gt \dfrac{4}{5\root{4}\of{5}}\)
- 2 when \(|k|=\dfrac{4}{5\root{4}\of{5}}\)
- 3 when \(|k| \lt \dfrac{4}{5\root{4}\of{5}}\)
3.6.7.14. (✳).
Answer.
(a)
(b) For any real \(x\text{,}\) define \(\sinh^{-1}(x)\) to be the unique solution of \(\sinh(y)=x\text{.}\) For every \(x\in[1,\infty)\text{,}\) define \(\cosh^{-1}(x)\) to be the unique \(y\in[0,\infty)\) that obeys \(\cosh(y)=x\text{.}\)
(c) \(\ds\diff{}{x}\{\cosh^{-1}(x)\}=\dfrac{1}{\sqrt{x^2-1}}\)
3.7 L’Hôpital’s Rule, Indeterminate Forms
3.7.4 Exercises
3.7.4.1.
Answer.
There are many possible answers. Here is one: \(f(x)=5x\text{,}\) \(g(x)=2x\text{.}\)
3.7.4.2.
Answer.
There are many possible answers. Here is one: \(f(x)=x\text{,}\) \(g(x)=x^2\text{.}\)
3.7.4.3.
Answer.
There are many possible answers. Here is one: \(f(x)=1+\frac{1}{x}\text{,}\) \(g(x)=x\log 5\) (recall we use \(\log\) to mean logarithm base \(e\)).
3.7.4.4. (✳).
Answer.
\(-\dfrac{2}{\pi}\)
3.7.4.5. (✳).
Answer.
\(-\infty\)
3.7.4.6. (✳).
Answer.
0
3.7.4.7. (✳).
Answer.
0
3.7.4.8. (✳).
Answer.
3
3.7.4.9.
Answer.
\(2\)
3.7.4.10. (✳).
Answer.
0
3.7.4.11. (✳).
Answer.
\(\frac{1}{2}\)
3.7.4.12.
Answer.
0
3.7.4.13.
Answer.
\(5\)
3.7.4.14.
Answer.
\(\infty\)
3.7.4.15. (✳).
Answer.
3
3.7.4.16. (✳).
Answer.
\(\frac{3}{2}\)
3.7.4.17. (✳).
Answer.
0
3.7.4.18. (✳).
Answer.
\(\frac{1}{3}\)
3.7.4.19.
Answer.
0
3.7.4.20.
Answer.
\(\frac{1}{\sqrt{e}}\)
3.7.4.21.
Answer.
1
3.7.4.22.
Answer.
1
3.7.4.23. (✳).
Answer.
\(c=0\)
3.7.4.24. (✳).
Answer.
\begin{equation*}
\lim\limits_{x\rightarrow 0}\dfrac{e^{k\sin(x^2)}-(1+2x^2)}{x^4}=\left\{\begin{array}{rl}
-\infty&k \lt 2\\
2&k=2\\
\infty&k \gt 2
\end{array}\right.
\end{equation*}
3.7.4.25.
Answer.
-
We want to find the limit as \(n\) goes to infinity of the percentage error, \(\ds\lim_{n \rightarrow \infty} 100\frac{|S(n)-A(n)|}{|S(n)|}\text{.}\) Since \(A(n)\) is a nicer function than \(S(n)\text{,}\) let’s simplify: \(\ds\lim_{n \rightarrow \infty} 100\frac{|S(n)-A(n)|}{|S(n)|} = 100\left|1-\ds\lim_{n \to \infty}\frac{A(n)}{S(n)}\right|\text{.}\)We figure out this limit the natural way:\begin{align*} 100\left|1-\ds\lim_{n \to \infty}\frac{A(n)}{S(n)}\right|&= 100\left|1-\ds\lim_{n \rightarrow \infty}\underbrace{\frac{5n^4}{5n^4-13n^3-4n+\log (n)}}_{\atp {\mathrm{num}\to\infty} {\mathrm{den}\to\infty}}\right|\\ &= 100\left|1-\ds\lim_{n \rightarrow \infty}\frac{20n^3}{20n^3-39n^2-4+\frac{1}{n}}\right|\\ &= 100\left|1-\ds\lim_{n \rightarrow \infty}\frac{n^3}{n^3}\cdot\frac{20}{20-\frac{39}{n}-\frac{4}{n^3}+\frac{1}{n^4}}\right|\\ &=100|1-1|=0 \end{align*}So, as \(n\) gets larger and larger, the relative error in the approximation gets closer and closer to 0.
- Now, let’s look at the absolute error.\begin{align*} \lim_{n \rightarrow \infty} \left| S(n)-A(n)\right|&=\lim_{n \rightarrow \infty} |-13n^3-4n+\log n|=\infty \end{align*}So although the error gets small relative to the giant numbers we’re talking about, the absolute error grows without bound.
4 Towards Integral Calculus
4.1 Introduction to Antiderivatives
4.1.2 Exercises
4.1.2.1.
Answer.
\(F(x)=f(x)+C\)
4.1.2.2.
Answer.
\(C(x)\)
4.1.2.3.
Answer.
\(F(x)=x^3+x^5+5x^2-9x+C\)
4.1.2.4.
Answer.
\(F(x)=\dfrac{3}{40}x^8-\dfrac{18}{5}x^5+\dfrac{1}{2}x^2+C\)
4.1.2.5.
Answer.
\(F(x)=3x^{\tfrac{4}{3}}+\dfrac{45}{17x^{1.7}}+C\)
4.1.2.6.
Answer.
\(F(x)=\dfrac{2}{7}\sqrt{x}+C\)
4.1.2.7.
Answer.
\(F(x)=\dfrac{1}{5}e^{5x+11}+C\)
4.1.2.8.
Answer.
\(F(x)=-\dfrac{3}{5}\cos(5x)+\dfrac{7}{13}\sin(13x)+C\)
4.1.2.9.
Answer.
\(F(x)=\tan(x+1)+C\)
4.1.2.10.
Answer.
\(F(x)=\log|x+2|+C\)
4.1.2.11.
Answer.
\(F(x)=\dfrac{7}{\sqrt{3}}\arcsin(x)+C\)
4.1.2.12.
Answer.
\(F(x)=\dfrac{1}{5}\arctan(5x)+C\)
4.1.2.13.
Answer.
\(f(x)=x^3-\dfrac{9}{2}x^2+4x+\dfrac{19}{2}\)
4.1.2.14.
Answer.
\(f(x)=\dfrac{1}{2}\sin(2x)+\pi\)
4.1.2.15.
Answer.
\(f(x)=\log|x|\)
4.1.2.16.
Answer.
\(f(x)=\arcsin x+x - \pi-1\)
4.1.2.17.
Answer.
It takes \(\frac{1}{2}\log 7\) hours (about 58 minutes) for the initial colony to increase by 300 individuals.
4.1.2.18.
Answer.
At time \(t\text{,}\) the amount of money in your bank account is \(75000e^{\tfrac{t}{50}}+C\) dollars, for some constant \(C\text{.}\)
4.1.2.19.
Answer.
\(\dfrac{24}{\pi}+6\approx13.6\) kWh
4.1.2.20. (✳).
Answer.
\(f'(x)=g'(x)=\dfrac{1}{\sqrt{x-x^2}}\text{;}\) \(f\) and \(g\) differ only by a constant.
4.1.2.21.
Answer.
\(F(x)=\sin(2x)\cos(3x)+C\)
4.1.2.22.
Answer.
\(F(x)=\dfrac{e^x}{x^2+1}+C\)
4.1.2.23.
Answer.
\(F(x)=e^{x^3}+C\)
4.1.2.24.
Answer.
\(F(x)=-\dfrac{5}{2}\cos(x^2)+C\)
4.1.2.25.
Answer.
\(F(x)=\dfrac{1}{2}x^2+C\text{.}\)
4.1.2.26.
Answer.
\(F(x)=7\arcsin\left(\dfrac{x}{\sqrt{3}}\right)+C\)
4.1.2.27.
Answer.
\(V(H)=2\pi\left(\dfrac{1}{5}H^5+\dfrac{2}{3}H^3+H\right)\)
4.1.2.28.
Answer.
\begin{equation*}
f(x)=\left\{\begin{array}{rl}
\log|x|&x \lt 0\\
\log x + C&x \gt 0
\end{array}\right.
\end{equation*}
where \(C\) is an arbitrary constant.
