# Rolle's theorem and mean value theorem 2 questions

How would I solve the following two questions.

Question 1

Determine whether the functions $$f(x)=3\sin(2x)$$ satisfies conditions of Rolle's theorem for the interval $$[0,\pi]$$. If so find all numbers $$c$$ that satisfy the conclusion of the theorem.

This be what I did.

I know $$f(x)=3\sin(2x)$$ is continuous on $$[0,\pi]$$ and differentiable on the interval $$(0,\pi)$$.

I know $$f'(c)=0$$ in Rolle's theorem.

So I did the derivative and got $$3\cos(2c)(2)$$=0

and then got $$6\cos2c=0$$ but how would I get $$c$$?

Question 2

Determine whether the function $$f(x)=1+\sqrt[3]{x^2}$$ satisfies conditions of the mean value theorem for the interval $$[-1,1]$$.

I know $$f'(x)=\frac{2}{3}x^{-1/3}$$ and for $$c$$ I got $$1$$ using $$\frac{f(b)-f(a)}{b-a}$$

then I set $$f'(x)=\frac{2}{3}x^{-1/3}=1$$ but how would I get $$c$$?

• For question 2 see this. Is there a $c\in[-1,1]: \ f'(c)=0$? Yet $f(1)=f(-1)$.
– P..
Commented Feb 21, 2013 at 21:59

Hint:

1. For $f(x)=3\sin(2x)$ is sufficient to solve equation $\cos{2c}=0 \Rightarrow 2c=\dfrac{\pi}{2} \Rightarrow c=\dfrac{\pi}{4}\;\;\;(c\in(0,\;\pi))$
2. Derivative of $f(x)=1+\sqrt[3]{x^2}$ is discontinuous at the point $x=0,$ so $f(x)=1+\sqrt[3]{x^2}$ does not satisfy theorem conditions.
• If it is discontinous then I cant find c? Commented Feb 21, 2013 at 21:42
• @FernandoMartinez You remember Rolle's theorem only works if the function is continuous on the closed interval. Think why? Commented Feb 21, 2014 at 9:22

I hope that for question 1, you also noted that $f(0)=f(\pi)$ as that is also part of the conditions. From $6\cos (2c)=0$ we see $\cos(2c)=0$, hence $2c\in\{\frac\pi2+k\pi\mid k\in\mathbb Z\}$ and $c\in\{\frac\pi4+k\frac\pi2\mid k\in\mathbb Z\}$. In order to have $c\in(0,\pi)$, we must (and can) take $k=0$ or $k=1$ i.e. the solutions are $c=\frac\pi4$ and $\frac{3\pi}4$.

For question 2, you didn't show whether or not the conditions of the MWT are met. Indeed, $f$ is continuous on $[a,b]=[0,1]$ and differentiable on $(a,b)$, so the MWT is applicable. You need to solve $$\frac23x^{-\frac13}=1\iff \frac23=x^{\frac13}\iff \frac8{27}=x.$$

EDIT: Oops, I treated $[a,b]=[0,1]$ in question 2, but the problem actually says $[a,b]=[-1,1]$. This means that we do not have a derivative $f'(x)$ for all $x\in(a,b)$, i.e. the conditions are not met.

• but my original function is continuous on the interval [-1,1] did I make a mistake? Commented Feb 21, 2013 at 21:54
• O i see never mind i get it. Commented Feb 21, 2013 at 21:59

Remark
In question 1 you didn't check one of the conditions of Rolle's Theorem, you have to make sure that $f(b)=f(a)$ for some $a,b\in [0,\pi]$ which in this case are $0,\pi$, now you can conclude that there is $c\in (0,\pi)$ such that $f'(c)=0$.

Finally, you have $6cos(2c)=0$ which implies tat $cos(2c)=0$ and we know when $cosx=0$ so it is easy to find the value of $c$.

First question:

$$6\cos 2c=0 \\ \cos 2c=0 \\ 2c=\cos ^{-1}0=\frac{\pi}{2}$$ Now solve for c.

Second question:

You made a mistake with $\frac{f(b)-f(a)}{b-a}=\frac{f(-1)-f(1)}{-1-1}=\frac{2-2}{-2}=0$. So, $$f'(x)=\frac{2}{3} x^{-\frac{1}{3}}=0$$

So, the derivative is discontinious at x=0 and the conditions of MWT are not met.