Science:Math Exam Resources/Courses/MATH110/December 2014/Question 03 (e)

MATH110 December 2014

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Question 03 (e)
Now consider the function $$g(x)={\begin{cases}e^{x}&{\text{if }}x\leq 0\\{\sqrt {x}}&{\text{if }}0<x<4\\\displaystyle mx+b&{\text{if }}x\geq 4.\end{cases}}$$ where ${\textstyle m}$ and ${\textstyle b}$ are constants. Find values for ${\textstyle m}$ and ${\textstyle b}$ such that ${\textstyle g(x)}$ is differentiable at ${\textstyle x=4.}$

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
Recall that if $g$ is not continuous at $x=a$, then it is not differentiable at the same point. Therefore, to be differentiable at $x=4$, $g$ must be continuous at the point first.

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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. As we mentioned in the Hint, $g$ must be continuous at $x=4$. This means $g(4)=\lim _{x\to 4^{+}}g(x)=\lim _{x\to 4^{-}}g(x)$. Since we have $g(4)=\lim _{x\to 4^{+}}g(x)=\lim _{x\to 4^{+}}(mx+b)=4m+b$ and $\lim _{x\to 4^{-}}g(x)={\sqrt {4}}=2$, for the continuity at $x=4$, we need $4m+b=2$. On the other hand, the slope of $g$ also should be matched. i.e., $\lim _{x\to 4^{+}}g'(x)=\lim _{x\to 4^{-}}g'(x)$. The left-hand side can be evaluated by $\lim _{x\to 4^{+}}g'(x)=\lim _{x\to 4^{+}}(mx+b)'=m$, while the right hand side becomes $\lim _{x\to 4^{-}}g'(x)=\lim _{x\to 4^{-}}({\sqrt {x}})'=\lim _{x\to 4^{-}}{\frac {1}{2{\sqrt {x}}}}={\frac {1}{2{\sqrt {4}}}}={\frac {1}{4}}$. Therefore, we get $m={\frac {1}{4}}$ and plugging this back to the equation $4m+b=2$, we obtain $b=1$. Answer: $\color {blue}m={\frac {1}{4}},b=1$