Science:Math Exam Resources/Courses/MATH110/December 2017/Question 07 (b)/Solution 1

At the points except for $x=-1$ and $x=0$, $h$ is differentiable. This is because polynomials and exponential functions are differentiable on the whole real line. Therefore,

$${\begin{aligned}h'(x)&={\begin{cases}(x^{2}-1)'&{\text{ if }}x<-1\\(x+1)'&{\text{ if }}-1<x<0\\(e^{x})'&{\text{ if }}x>0\end{cases}}\\&={\begin{cases}2x&{\text{ if }}x<-1\\1&{\text{ if }}-1<x<0\\e^{x}&{\text{ if }}x>0.\end{cases}}\end{aligned}}$$

As mentioned in the hint, we use the limit definition of a derivative to get $h'(-1)$ and $h'(0)$.

First, to get $h'(-1)$, we consider

$$\lim _{x\to -1+}{\frac {h(x)-h(-1)}{x-(-1)}}=\lim _{x\to -1+}{\frac {x+1-0}{x+1}}=1,$$

$$\lim _{x\to -1-}{\frac {h(x)-h(-1)}{x-(-1)}}=\lim _{x\to -1-}{\frac {x^{2}-1-0}{x+1}}==\lim _{x\to -1-}(x-1)=-2.$$

Therefore, the limit doesn't exist, so that $h'(-1)$ also doesn't exists.

On the other hand, to find $h'(0)$, we consider

$$\lim _{x\to 0+}{\frac {h(x)-h(0)}{x-0}}=\lim _{x\to 0+}{\frac {e^{x}-1}{x}}=(e^{x})'|_{x=0}=e^{x}|_{x=0}=1$$

$$\lim _{x\to 0-}{\frac {h(x)-h(0)}{x-0}}=\lim _{x\to 0-}{\frac {x+1-1}{x-0}}=1.$$

Therefore, the derivative of $h$ exists at $x=0$ with the value

$$h'(0)=\lim _{x\to 0}{\frac {h(x)-h(0)}{x-0}}=1.$$

Based on this analysis, we can draw the graph of $h'$ as follows.

Apparently, there's no $x$-intercept and the only $y$-intercept is $y=1$ indicated by the red dot on the graph.