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

MATH110 December 2017

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Question 07 (b)
(b) Sketch the graph of the derivative $h'(x)$. Make sure you identify all $x$- and $y$-intercepts (if they exist) by writing down the coordinates of those points on your graph. Also identify any $x$ value for which $h'(x)$ does not exist.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the definition of the derivative to get $h'(0)$ and $h'(-1)$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.