Science:Math Exam Resources/Courses/MATH110/December 2015/Question 03 (c)

MATH110 December 2015

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Question 03 (c)
Consider the following function, $f(x)={\begin{cases}e^{x}&x\leq 0\\ax+b&0<x<1\\-x^{2}+x&x\geq 1.\end{cases}}$ where $a$ and $b$ are constants. (c) At which point(s) is the function $f$ drawn above differentiable? Justify your answer.

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
Check differentiability for the points where the pieces meet each other, i.e. at $x=0$ and $x=1$ .

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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. We recall that a function $f$ is NOT differentiable at a point $x$ if the limit $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ doesn't exist. Also, if a function has a corner, a cusp or a vertical tangent line for the graph, at that point the function is NOT differentiable. Obviously from the graph of $f$ in part (b), at $x=0$, we have a corner. Therefore, $f$ is not differentiable at this point. To check this by using the definition, we first calculate the left hand limit $\lim _{h\to 0-}{\frac {f(h)-f(0)}{h}}=\lim _{h\to 0-}{\frac {e^{h}-e^{0}}{h}}=(e^{x})'{\bigg |}_{x=0}=1$. The first equality follows from that we consider negative $h$ and for such $h$, $f(h)=e^{h}$. However, we have the right hand limit $\lim _{h\to 0+}{\frac {f(h)-f(0)}{h}}=\lim _{h\to 0+}{\frac {-h+1-1}{h}}=-1$. Since the left hand limit doesn't match with the right hand one, the limit doesn't exist. At $x=1$, we have the left hand limit $\lim _{h\to 0-}{\frac {f(1+h)-f(1)}{h}}=\lim _{h\to 0-}{\frac {-(1+h)+1}{h}}=-1$ and the right hand limit $\lim _{h\to 0+}{\frac {f(1+h)-f(1)}{h}}=(-x^{2}+x)'|_{x=1}=(-2x+1)|_{x=1}=-1$. Therefore, the function is differentiable at $x=1$. Everywhere else, there is no corner, cusp or a vertical tangent line for the graph, so the function is differentiable everywhere except at $x=0$, i.e. on $\color {blue}\mathbb {R} \setminus \{0\}$