Science:Math Exam Resources/Courses/MATH100/December 2015/Question 12 (a)

MATH100 December 2015

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Question 12 (a)
Let $g(x)$ be a continuous function for which $g'(x)$ and $g''(x)$ exist. If $g(x)$ has at least three zeroes, then how many zeroes must $g'(x)$ and $g''(x)$ have? Explain your answer carefully.

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
Apply Mean Value Theorem to $g$ and $g'$.

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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 mean value theorem: if $f$ is differentiable on $[a_{0},b_{0}]$, then there exists a number $c_{0}\in (a_{0},b_{0})$ such that $f'(c_{0})={\frac {f(b_{0})-f(a_{0})}{b_{0}-a_{0}}}.$ Let $a<b<c$ be the three zeros of the function $g$. i.e., $g(a)=g(b)=g(c)=0$. Note that $g$ and $g'$ are differentiable, so that we can apply this theorem for these functions. First, apply the mean value theorem for $f=g$, $a_{0}=a$, and $b_{0}=b$, then we can find $d\in (a,b)$ such that $g'(d)={\frac {g(b)-g(a)}{b-a}}=0.$ Similarly, applying the theorem for $f=g$, $a_{0}=b$, and $b_{0}=c$, we can find $e\in (b,c)$ such that $g'(e)={\frac {g(c)-g(b)}{c-b}}=0.$ Therefore, $g'$ must have at least two zeros. Now, we apply the mean value theorem for $f=g'$, $a_{0}=d$ and $b_{0}=e$, then we get $x\in (d,e)$ such that $g''(x)=(g')'(x)={\frac {g'(e)-g'(d)}{e-d}}=0.$ Note that since $a<d<b<e<c$, $d\neq e$, As a result, $g''$ must have at least one zero. Therefore, the answer is $\color {blue}{2,1}$.