Science:Math Exam Resources/Courses/MATH110/April 2017/Question 07 (c)

MATH110 April 2017

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Question 07 (c)
For the value of $k$ found in part (b), prove that there exists a number $c$ in $(-4,2)$ such that $f'(c)=-1$. Make sure you justify your claims.

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
Since we have solved ${\textstyle k}$ in (b), all we need to do is check the conditions in the mean value theorem stated in (a): continuity on ${\textstyle [-4,2]}$ and differentiability on ${\textstyle (-4,2)}$, then we can apply the theorem. (Note that left derivative equals right derivative at that point means differentiability at that 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. First let's write down the equation $$f(x)={\begin{cases}-3x+2,\ \ \ \ \ \ \ \ \ \ \ \ -4\leq x\leq 0\\x^{3}+x^{2}-3x+2,\ \ 0<x\leq 2\end{cases}}$$ (1) For continuity, we already knew from (b) that ${\textstyle f}$ is continuous on the entire interval ${\textstyle [-4,2]}$ (2) For differentiability, ${\textstyle -3x+2}$ is differentiable on ${\textstyle (-4,0)}$, ${\textstyle x^{3}+x^{2}-3x+2}$ is differentiable on ${\textstyle (0,2)}$ since they are both polynomials on the domain, but we have to check whether ${\textstyle f}$ is differentiable at point ${\textstyle x=0}$. (we only know it is continuous at this point for now.) So we need to find the left derivative and right derivative, they are $$f'(0^{-})=\lim _{h\rightarrow 0^{-}}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0^{-}}{\frac {-3h}{h}}=-3,$$ and $$f'(0^{+})=\lim _{h\rightarrow 0^{+}}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0^{+}}{\frac {h^{3}+h^{2}-3h}{h}}=-3.$$ Then we know ${\textstyle f}$ is also differentiable at ${\textstyle x=0}$, thus by definition of differentiability ${\textstyle f}$ is differentiable on the interval ${\textstyle (-4,2)}$. Now if we apply the mean value theorem for ${\textstyle a=-4,b=2}$, then there exists ${\textstyle c\in (-4,2)}$ such that $$f'(c)={\frac {f(b)-f(a)}{b-a}}={\frac {f(2)-f(-4)}{2-(-4)}}={\frac {8-14}{6}}=-1.$$