Science:Math Exam Resources/Courses/MATH104/December 2016/Question 09 (b)

MATH104 December 2016

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Question 09 (b)
Consider the following function; $f(x)={\begin{cases}e^{x},&x\leq 0\\ax+b&0<x\leq 1\\1+{\sqrt {x}}&x>1.\end{cases}}$ (b) Determine whether or not $f(x)$ can be differentiable at $x=0$ or $x=1$.

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?

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Hint
Use the definition of the differentiability of a function at some point.

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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. Recall the a function $f$ is differentiable at $x=a$ if $\lim _{x\to a^{-}}{\frac {f(x)-f(a)}{x-a}}=\lim _{x\to a^{+}}{\frac {f(x)-f(a)}{x-a}}$. Before we start, note that we need $f(x)$ to be continuous in order to be differentiable at $x=0$ or $x=1$. So, from the part (a) we know that $a=b=1.$ Now, we check the differentiability of the given function $f$ at $x=0$. Evaluating the limits, we get $\lim _{x\to 0^{-}}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0^{-}}{\frac {e^{x}-1}{x}}=(e^{x})'{\bigg |}_{x=0}=e^{x}{\bigg |}_{x=0}=1.$ and $\lim _{x\to 0^{+}}{\frac {f(x)-f(0)}{x-0}}=\lim _{x\to 0^{-}}{\frac {x+1-1}{x}}=1,$ Therefore, by the definition above, $f$ is differentiable at $x=0$. However, we have $\lim _{x\to 1^{-}}{\frac {f(x)-f(1)}{x-1}}=\lim _{x\to 1^{-}}{\frac {x+1-2}{x-1}}=1$ but $\lim _{x\to 1^{+}}{\frac {f(x)-f(1)}{x-1}}=\lim _{x\to 1^{+}}{\frac {1+{\sqrt {x}}-2}{x-1}}=\lim _{x\to 1^{+}}{\frac {{\sqrt {x}}-1}{({\sqrt {x}}-1)({\sqrt {x}}+1)}}={\frac {1}{2}}\neq 1$. This implies that $f$ is not differentiable at $x=1$. To sum $\color {blue}f'(0)=1,f'(1):DNE$.