CLP-1 Differential Calculus

Let \(f(x) = x^2\) and \(g(x)=x^3\text{.}\) Then

Hence \(f(x)\) is even and \(g(x)\) is odd.

Notice any polynomial involving only even powers of \(x\) will be even

Similarly any polynomial involving only odd powers of \(x\) will be odd

Consider the function

• The function is even since
  \begin{align*} g(-x) &= \frac{(-x)^2-9}{(-x)^2+3} = \frac{x^2-9}{x^2+3} = g(x) \end{align*}
  Thus it suffices to study the function for \(x\geq0\) because we can then use the even symmetry to understand what happens for \(x \lt 0\text{.}\)
• The function is defined on all real numbers since its denominator \(x^2+3\) is never zero. Hence it has no vertical asymptotes.
• The \(y\)-intercept is \(g(0) = \frac{-9}{3} = -3\text{.}\) And \(x\)-intercepts are given by the solution of \(x^2-9=0\text{,}\) namely \(x=\pm 3\text{.}\) Note that we only need to establish \(x=3\) as an intercept. Then since \(g\) is even, we know that \(x=-3\) is also an intercept.
• To find the horizontal asymptotes we compute the limit as \(x\to+\infty\)
  \begin{align*} \lim_{x\to \infty} g(x) &= \lim_{x\to \infty} \frac{x^2-9}{x^2+3}\\ &= \lim_{x\to \infty} \frac{x^2(1-9/x^2)}{x^2(1+3/x^2)}\\ &= \lim_{x\to \infty} \frac{1-9/x^2}{1+3/x^2} = 1 \end{align*}
  Thus \(y=1\) is a horizontal asymptote. Indeed, this is also the asymptote as \(x\to-\infty\) since by the even symmetry
  \begin{align*} \lim_{x\to -\infty} g(x) &=\lim_{x\to \infty} g(-x) = \lim_{x\to \infty} g(x). \end{align*}

• We can already produce a quite reasonable sketch just by putting in the horizontal asymptote and the intercepts and drawing a smooth curve between them.

  

  Note that we have drawn the function as never crossing the asymptote \(y=1\text{,}\) however we have not yet proved that. We could by trying to solve \(g(x)=1\text{.}\)

  \begin{align*} \frac{x^2-9}{x^2+3} &= 1\\ x^2-9 &= x^2+3\\ -9=3 & \text{ so no solutions.} \end{align*}

  Alternatively we could analyse the first derivative to see how the function approaches the asymptote.

• Now we turn to the first derivative:
  \begin{align*} g'(x) &= \frac{(x^2+3)(2x) - (x^2-9)(2x)}{(x^2+3)^2}\\ &= \frac{24x}{(x^2+3)^2} \end{align*}
  There are no singular points since the denominator is nowhere zero. The only critical point is at \(x=0\text{.}\) Thus we must find the sign of \(g'(x)\) on the intervals
  \begin{align*} (-\infty,0) && (0,\infty) \end{align*}
• When \(x \gt 0\text{,}\) \(24x \gt 0\) and \((x^2+3) \gt 0\text{,}\) so \(g'(x) \gt 0\) and the function is increasing. By even symmetry we know that when \(x \lt 0\) the function must be decreasing. Hence the critical point \(x=0\) is a local minimum of the function.
• Notice that since the function is increasing for \(x \gt 0\) and the function must approach the horizontal asymptote \(y=1\) from below. Thus the sketch above is quite accurate.
• Now consider the second derivative:
  \begin{align*} g''(x) &= \diff{}{x} \frac{24x}{(x^2+3)^2}\\ &= \frac{(x^2+3)^2 \cdot 24 - 24x\cdot 2 (x^2+3)\cdot2x}{(x^2+3)^4}\\ \end{align*}

  cancel a factor of \((x^2+3)\)

  \begin{align*} &= \frac{(x^2+3) \cdot 24 - 96x^2}{(x^2+3)^3}\\ &= \frac{72(1-x^2)}{(x^2+3)^3} \end{align*}
• It is clear that \(g''(x) = 0\) when \(x=\pm 1\text{.}\) Note that, again, we can infer the zero at \(x=-1\) from the zero at \(x=1\) by the even symmetry. Thus we need to examine the sign of \(g''(x)\) the intervals
  \begin{align*} (-\infty,-1)&&(-1,1)&&(1,\infty) \end{align*}
• When \(|x| \lt 1\) we have \((1-x^2) \gt 0\) so that \(g''(x) \gt 0\) and the function is concave up. When \(|x| \gt 1\) we have \((1-x^2) \lt 0\) so that \(g''(x) \lt 0\) and the function is concave down. Thus the points \(x=\pm 1\) are inflection points. Their coordinates are \((\pm1, g(\pm1)) =(\pm 1,-2)\text{.}\)

• Putting this together gives the following sketch:

  

The classic example of a periodic function is \(f(x)=\sin x\text{,}\) which has period \(2\pi\) since \(f(x+2\pi)=\sin(x+2\pi)=\sin x=f(x)\text{.}\)