CLP-1 Differential Calculus

1. Reading from \(f(x)\text{:}\)

   • The function is a polynomial so it is defined everywhere.
   • Since \(f(-x) = -x^3+3x+1 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
   • The \(y\)-intercept is \(y=1\text{.}\) The \(x\)-intercepts are not easily computed since it is a cubic polynomial that does not factor nicely ²With the aid of a computer we can find the \(x\)-intercepts numerically: \(x\approx -1.879385242, 0.3472963553\text{,}\) and \(1.532088886\text{.}\) If you are interested in more details check out Appendix C. . So for this example we don't worry about finding them.
   • Since it is a polynomial it has no vertical asymptotes.
   • For very large \(x\text{,}\) both positive and negative, the \(x^3\) term in \(f(x)\) dominates the other two terms so that
     \begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ -\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}
     and there are no horizontal asymptotes.

2. We now compute the derivative:

   \begin{align*} f'(x) &= 3x^2-3 = 3(x^2-1)=3(x+1)(x-1) \end{align*}
   • The critical points (where \(f'(x)=0\)) are at \(x=\pm 1\text{.}\) Further since the derivative is a polynomial it is defined everywhere and there are no singular points. The critical points split the real line into the intervals \((-\infty,-1),(-1,1)\) and \((1,\infty)\text{.}\)
   • When \(x \lt -1\text{,}\) both factors \((x+1),(x-1) \lt 0\) so \(f'(x) \gt 0\text{.}\)
   • Similarly when \(x \gt 1\text{,}\) both factors \((x+1),(x-1) \gt 0\) so \(f'(x) \gt 0\text{.}\)
   • When \(-1 \lt x \lt 1\text{,}\) \((x-1) \lt 0\) but \((x+1) \gt 0\) so \(f'(x) \lt 0\text{.}\)

   • Summarising all this

     	\((-\infty,-1)\)	-1	(-1,1)	1	\((1,\infty)\)
     \(f'(x)\)	positive	0	negative	0	positive
     	increasing	maximum	decreasing	minimum	increasing

     So \((-1,f(-1))=(-1,3)\) is a local maximum and \((1,f(1))=(1,-1)\) is a local minimum.

3. Compute the second derivative:

   \begin{gather*} f''(x) = 6x \end{gather*}
   • The second derivative is zero when \(x=0\text{,}\) and the problem is quite easy to analyse. Clearly, \(f''(x) \lt 0\) when \(x \lt 0\) and \(f''(x) \gt 0\) when \(x \gt 0\text{.}\)
   • Thus \(f\) is concave down for \(x \lt 0\text{,}\) concave up for \(x \gt 0\) and has an inflection point at \(x=0\text{.}\)

Putting this all together gives:

1. Reading from \(f(x)\text{:}\)

   • The function is a polynomial so it is defined everywhere.
   • Since \(f(-x) = x^4+4x^3 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
   • The \(y\)-intercept is \(y=f(0)=0\text{,}\) while the \(x\)-intercepts are given by the solution of
     \begin{align*} f(x)=x^4-4x^3 &= 0\\ x^3(x-4)&=0 \end{align*}
     Hence the \(x\)-intercepts are \(0,4\text{.}\)
   • Since \(f\) is a polynomial it does not have any vertical asymptotes.
   • For very large \(x\text{,}\) both positive and negative, the \(x^4\) term in \(f(x)\) dominates the other term so that
     \begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ +\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}
     and the function has no horizontal asymptotes.

2. Now compute the derivative \(f'(x)\text{:}\)

   \begin{gather*} f'(x) = 4x^3-12x^2 = 4(x-3)x^2 \end{gather*}
   • The critical points are at \(x=0,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,0)\text{,}\) \((0,3)\) and \((3,\infty)\text{.}\)
   • When \(x \lt 0\text{,}\) \(x^2 \gt 0\) and \(x-3 \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
   • When \(0 \lt x \lt 3\text{,}\) \(x^2 \gt 0\) and \(x-3 \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
   • When \(3 \lt x\text{,}\) \(x^2 \gt 0\) and \(x-3 \gt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)

   • Summarising all this

     	\((-\infty,0)\)	0	(0,3)	3	\((3,\infty)\)
     \(f'(x)\)	negative	0	negative	0	positive
     	decreasing	horizontal tangent	decreasing	minimum	increasing

     So the point \((3,f(3))=(3,-27)\) is a local minimum. The point \((0,f(0))=(0,0)\) is neither a minimum nor a maximum, even though \(f'(0)=0\text{.}\)

3. Now examine \(f''(x)\text{:}\)

   \begin{gather*} f''(x) = 12x^2-24x=12x(x-2) \end{gather*}
   • So \(f''(x)=0\) when \(x=0,2\text{.}\) This splits the real line into the intervals \((-\infty,0),(0,2)\) and \((2,\infty)\text{.}\)
   • When \(x \lt 0\text{,}\) \(x-2 \lt 0\) and so \(f''(x) \gt 0\text{.}\)
   • When \(0 \lt x \lt 2\text{,}\) \(x \gt 0\) and \(x-2 \lt 0\) and so \(f''(x) \lt 0\text{.}\)
   • When \(2 \lt x\text{,}\) \(x \gt 0\) and \(x-2 \gt 0\) and so \(f''(x) \gt 0\text{.}\)
   • Thus the function is convex up for \(x \lt 0\text{,}\) then convex down for \(0 \lt x \lt 2\text{,}\) and finally convex up again for \(x \gt 2\text{.}\) Hence \((0,f(0))=(0,0)\) and \((2,f(2))=(2,-16)\) are inflection points.

Putting all this information together gives us the following sketch.

1. Reading from \(f(x)\text{:}\)

   • The function is a polynomial so it is defined everywhere.
   • Since \(f(-x) = -x^3-6x^2-9x-54 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
   • The \(y\)-intercept is \(y=f(0)=-54\text{,}\) while the \(x\)-intercepts are given by the solution of
     \begin{align*} f(x)=x^3-6x^2+9x-54 &= 0\\ x^2(x-6) + 9(x-6) &=0\\ (x^2+9)(x-6) &= 0 \end{align*}
     Hence the only \(x\)-intercept is \(6\text{.}\)
   • Since \(f\) is a polynomial it does not have any vertical asymptotes.
   • For very large \(x\text{,}\) both positive and negative, the \(x^3\) term in \(f(x)\) dominates the other term so that
     \begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ -\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}
     and the function has no horizontal asymptotes.

2. Now compute the derivative \(f'(x)\text{:}\)

   \begin{align*} f'(x) &= 3x^2-12x+9\\ &= 3(x^2-4x+3) = 3(x-3)(x-1) \end{align*}
   • The critical points are at \(x=1,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,1)\text{,}\) \((1,3)\) and \((3,\infty)\text{.}\)
   • When \(x \lt 1\text{,}\) \((x-1) \lt 0\) and \((x-3) \lt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)
   • When \(1 \lt x \lt 3\text{,}\) \((x-1) \gt 0\) and \((x-3) \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
   • When \(3 \lt x\text{,}\) \((x-1) \gt 0\) and \((x-3) \gt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)

   • Summarising all this

     	\((-\infty,1)\)	1	(1,3)	3	\((3,\infty)\)
     \(f'(x)\)	positive	0	negative	0	positive
     	increasing	maximum	decreasing	minimum	increasing

     So the point \((1,f(1))=(1,-50)\) is a local maximum. The point \((3,f(3))=(3,-54)\) is a local minimum.

3. Now examine \(f''(x)\text{:}\)

   \begin{gather*} f''(x) = 6x-12 \end{gather*}
   • So \(f''(x)=0\) when \(x=2\text{.}\) This splits the real line into the intervals \((-\infty,2)\) and \((2,\infty)\text{.}\)
   • When \(x \lt 2\text{,}\) \(f''(x) \lt 0\text{.}\)
   • When \(x \gt 2\text{,}\) \(f''(x) \gt 0\text{.}\)
   • Thus the function is convex down for \(x \lt 2\text{,}\) then convex up for \(x \gt 2\text{.}\) Hence \((2,f(2))=(2,-52)\) is an inflection point.

Putting all this information together gives us the following sketch.

and if we zoom in around the interesting points (minimum, maximum and inflection point), we have

1. Reading from \(f(x)\text{:}\)

   • The function is rational so it is defined except where its denominator is zero — namely at \(x=\pm2\text{.}\)
   • Since \(f(-x) = \dfrac{-x}{x^2-4} = - f(x)\text{,}\) it is odd. Indeed this means that we only need to examine what happens to the function for \(x \geq 0\) and we can then infer what happens for \(x\leq 0\) using \(f(-x) = -f(x)\text{.}\) In practice we will sketch the graph for \(x\geq0\) and then infer the rest from this symmetry.
   • The \(y\)-intercept is \(y=f(0)=0\text{,}\) while the \(x\)-intercepts are given by the solution of \(f(x)=0\text{.}\) So the only \(x\)-intercept is \(0\text{.}\)
   • Since \(f\) is rational, it may have vertical asymptotes where its denominator is zero — at \(x=\pm 2\text{.}\) Since the function is odd, we only have to analyse the asymptote at \(x=2\) and we can then infer what happens at \(x=-2\) by symmetry.
     \begin{align*} \lim_{x\to 2^+} f(x) &= \lim_{x\to 2^+} \frac{x}{(x-2)(x+2)} = + \infty\\ \lim_{x\to 2^-} f(x) &= \lim_{x\to 2^-} \frac{x}{(x-2)(x+2)} = - \infty \end{align*}
   • We now check for horizontal asymptotes:
     \begin{align*} \lim_{x\to +\infty} f(x) &= \lim_{x\to +\infty} \frac{x}{x^2-4}\\ &= \lim_{x\to +\infty} \frac{1}{x-4/x} = 0 \end{align*}

2. Now compute the derivative \(f'(x)\text{:}\)

   \begin{align*} f'(x) &= \frac{(x^2-4)\cdot 1 - x\cdot 2x}{(x^2-4)^2}\\ &= \frac{-(x^2+4)}{(x^2-4)^2} \end{align*}
   • Hence there are no critical points. There are singular points where the denominator is zero, namely \(x=\pm2\text{.}\) Before we proceed, notice that the numerator is always negative and the denominator is always positive. Hence \(f'(x) \lt 0\) except at \(x=\pm 2\) where it is undefined.
   • The function is decreasing except at \(x=\pm 2\text{.}\)
   • We already know that at \(x = 2\) we have a vertical asymptote and that \(f'(x) \lt 0\) for all \(x\text{.}\) So
     \begin{gather*} \lim_{x\rightarrow 2} f'(x) = -\infty \end{gather*}

   • Summarising all this

     	[0,2)	2	\((2,\infty)\)
     \(f'(x)\)	negative	DNE	negative
     	decreasing	vertical asymptote	decreasing

     Remember — we will draw the graph for \(x\geq 0\) and then use the odd symmetry to infer the graph for \(x \lt 0\text{.}\)

3. Now examine \(f''(x)\text{:}\)

   \begin{align*} f''(x) &=- \frac{(x^2-4)^2\cdot(2x) - (x^2+4)\cdot2\cdot 2x\cdot(x^2-4)}{(x^2-4)^4}\\ &=- \frac{(x^2-4)\cdot(2x) - (x^2+4)\cdot4x}{(x^2-4)^3}\\ &=- \frac{2x^3-8x - 4x^3-16x}{(x^2-4)^3}\\ &= \frac{2x(x^2+12)}{(x^2-4)^3} \end{align*}
   • So \(f''(x)=0\) when \(x=0\) and does not exist when \(x=\pm 2\text{.}\) This splits the real line into the intervals \((-\infty,-2), (-2,0), (0,2)\) and \((2,\infty)\text{.}\) However we only need to consider \(x \geq 0\) (because of the odd symmetry).
   • When \(0 \lt x \lt 2\text{,}\) \(x \gt 0, (x^2+12) \gt 0\) and \((x^2-4) \lt 0\) so \(f''(x) \lt 0\text{.}\)
   • When \(x \gt 2\text{,}\) \(x \gt 0, (x^2+12) \gt 0\) and \((x^2-4) \gt 0\) so \(f''(x) \gt 0\text{.}\)

Putting all this information together gives the following sketch for \(x \geq 0\text{:}\)

We can then draw in the graph for \(x \lt 0\) using \(f(-x) = -f(x)\text{:}\)

Notice that this means that the concavity changes at \(x=0\text{,}\) so the point \((0,f(0))=(0,0)\) is a point of inflection (as indicated).

1. Reading from \(f(x)\text{:}\)

   • First notice that we can rewrite
     \begin{align*} f(x) &= \root{3}\of{\frac{x^2}{(x-6)^2}} = \root{3}\of{\frac{x^2}{x^2\cdot(1-6/x)^2}} = \root{3}\of{\frac{1}{(1-6/x)^2}} \end{align*}
   • The function is the cube root of a rational function. The rational function is defined except at \(x=6\text{,}\) so the domain of \(f\) is all reals except \(x=6\text{.}\)
   • Clearly the function is not periodic, and examining
     \begin{align*} f(-x) &= \root{3}\of{\frac{ 1}{(1-6/(-x))^2}}\\ &= \root{3}\of{\frac{1}{(1+6/x)^2}} \neq \pm f(x) \end{align*}
     shows the function is neither even nor odd.
   • To compute horizontal asymptotes we examine the limit of the portion of the function inside the cube-root
     \begin{gather*} \lim_{x\rightarrow\pm\infty} \frac{1}{(1-\frac{6}{x})^2} =1 \end{gather*}
     This means we have
     \begin{gather*} \lim_{x\rightarrow\pm\infty} f(x)=1 \end{gather*}
     That is, the line \(y=1\) will be a horizontal asymptote to the graph \(y=f(x)\) both for \(x\rightarrow+\infty\) and for \(x\rightarrow-\infty\text{.}\)
   • Our function \(f(x)\rightarrow+\infty\) as \(x\rightarrow 6\text{,}\) because of the \((1-6/x)^2\) in its denominator. So \(y=f(x)\) has \(x=6\) as a vertical asymptote.

2. Now compute \(f'(x)\text{.}\) Since we rewrote

   \begin{align*} f(x) &= \root{3}\of{\frac{1}{(1-6/x)^2}} =\left(1-\frac{6}{x}\right)^{-\frac{2}{3}} \end{align*}

   we can use the chain rule

   \begin{gather*} f'(x) = -\frac{2}{3}{\left(1-\frac{6}{x}\right)}^{-\frac{5}{3}}\frac{6}{x^2}\\ =-4 {\left(\frac{x-6}{x}\right)}^{-\frac{5}{3}}\frac{1}{x^2}\\ =-4 {\left(\frac{1}{x-6}\right)}^{\frac{5}{3}}\frac{1}{x^{\frac{1}{3}}} \end{gather*}
   • Notice that the derivative is nowhere equal to zero, so the function has no critical points. However there are two places the derivative is undefined. The terms
     \begin{align*} \left(\frac{1}{x-6}\right)^{\frac{5}{3}} && \frac{1}{x^{\frac{1}{3}}} \end{align*}
     are undefined at \(x=6,0\) respectively. Hence \(x=0,6\) are singular points. These split the real line into the intervals \((-\infty,0), (0,6)\) and \((6,\infty)\text{.}\)
   • When \(x \lt 0\text{,}\) \((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \lt 0\) and so \(f'(x)=-4 \cdot (\text{negative})\cdot(\text{negative}) \lt 0 \text{.}\)
   • When \(0 \lt x \lt 6\text{,}\) \((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \gt 0\) and so \(f'(x) \gt 0\text{.}\)
   • When \(x \gt 6\text{,}\) \((x-6) \gt 0\text{,}\) we have that \((x-6)^{-\frac53} \gt 0\) and \(x^{-\frac13} \gt 0\) and so \(f'(x) \lt 0\text{.}\)
   • We should also examine the behaviour of the derivative as \(x \to 0\) and \(x\to 6\text{.}\)
     \begin{align*} \lim_{x \to 0^-} f'(x) &= -4 \left( \lim_{x \to 0^-} (x-6)^{-\frac53} \right) \left( \lim_{x \to 0^-} x^{-\frac13} \right) = -\infty\\ \lim_{x \to 0^+} f'(x) &= -4 \left( \lim_{x \to 0^+} (x-6)^{-\frac53} \right) \left( \lim_{x \to 0^+} x^{-\frac13} \right) = +\infty\\ \lim_{x \to 6^-} f'(x) &= -4 \left( \lim_{x \to 6^-} (x-6)^{-\frac53} \right) \left( \lim_{x \to 6^-} x^{-\frac13} \right) = +\infty\\ \lim_{x \to 6^+} f'(x) &= -4 \left( \lim_{x \to 6^+} (x-6)^{-\frac53} \right) \left( \lim_{x \to 6^+} x^{-\frac13} \right) = -\infty \end{align*}
     We already know that \(x=6\) is a vertical asymptote of the function, so it is not surprising that the lines tangent to the graph become vertical as we approach 6. The behavior around \(x=0\) is less standard, since the lines tangent to the graph become vertical, but \(x=0\) is not a vertical asymptote of the function. Indeed the function takes a finite value \(y=f(0)=0\text{.}\)

   • Summarising all this

     	\((-\infty,0)\)	0	(0,6)	6	\((6,\infty)\)
     \(f'(x)\)	negative	DNE	positive	DNE	negative
     	decreasing	vertical tangents	increasing	vertical asymptote	decreasing

3. Now look at \(f''(x)\text{:}\)

   \begin{align*} f''(x)&=-4\diff{}{x} \left[{\left(\frac{1}{x-6}\right)}^{\frac{5}{3}} \frac{1}{x^{\frac{1}{3}}} \right]\\ \amp=-4 \left[-\frac{5}{3}{\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{1}{3}}} -\frac{1}{3} {\left(\frac{1}{x-6}\right)}^{\frac{5}{3}} \frac{1}{x^{\frac{4}{3}}}\right]\\ &=\frac{4}{3} {\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{4}{3}}}\ \left[5x +(x-6)\right]\\ &=8 {\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{4}{3}}}\ \left[x-1\right] \end{align*}

   Oof!

   • Both of the factors \({\Big(\frac{1}{x-6}\Big)}^{\frac{8}{3}} ={\Big(\frac{1}{\root{3}\of{x-6}}\Big)}^8\) and \(\frac{1}{x^{\frac{4}{3}}} =\Big(\frac{1}{\root{3}\of{x}}\Big)^4\) are even powers and so are positive (though possibly infinite). So the sign of \(f''(x)\) is the same as the sign of the factor \(x-1\text{.}\) Thus

     	\((-\infty,1)\)	1	\((1,\infty)\)
     \(f''(x)\)	negative	0	positive
     	concave down	inflection point	concave up

Here is a sketch of the graph \(y=f(x)\text{.}\)

It is hard to see the inflection point at \(x=1\text{,}\) \(y=f(1)=\frac{1}{ \root{3}\of{25} }\) in the above sketch. So here is a blow up of the part of the sketch around \(x=1\text{.}\)

And if we zoom in even more we have