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I am confused as to what my book intends for me to do here. It is asking me to let $f(x)= 1-x^{2/3}$ and show that $f(-1)=f(1)$ but there there is no number $c$ in $(-1,1)$ such that the derivative is equal to $0$. Also why does this not contradict Rolle's Theorem?

I am getting stuck on finding a way to make those two functions equal.

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The derivatives are equal, but that does not make the functions equal just the tangent line at points 1 and -1 I believe. – user138246 Oct 7 '11 at 21:30
Rolle's theorem requires derivability on $(-1,1)$. Did you check if it's the case? – Davide Giraudo Oct 7 '11 at 21:32
$f(1) = 1 - 1^{2/3} = 1-1 = 0$ and $f(-1) = 1 - (-1)^{2/3} = 1-1^{1/3} = 0$. I can't see your problem here. – martini Oct 7 '11 at 21:32
@Jordan: The question is: does $f$ have a derivative defined everywhere on $(-1,1)$? What is the derivative of $f$? You can probably figure that one out easily enough. Now look at the formula you got. Does that formula make sense for every number between $-1$ and $1$? – Arturo Magidin Oct 7 '11 at 21:36
"[...] make those two functions equal." This line of text shows that there is something you have gravely misunderstood. The value of a function (from $\mathbb{R}$ to $\mathbb{R}$) at a point is not itself a function. I did not understand at all what you were talking about until I read through the comments and realised that what you were having problems with was verifying that $f(-1) = f(1)$. – kahen Oct 7 '11 at 22:04

First, you probably intended to write $f(x) = 1-(x^2)^{1/3}$, so that $f(-x) = f(x)$, $\forall x \in \mathbb{R}$.

You already computed the derivative, but notice that there is a point where the derivative is not defined. Indeed, computing the derivative using the chain rule: $$ f^\prime(x) = \left( \frac{1}{3} (x^2)^{-2/3} \right) \cdot ( 2 x) = \frac{2}{3} \frac{x}{(x^2)^{2/3}} $$ Notice how $f^\prime(x) < 0$ for $x <0 $ and $f^\prime(x) > 0$ for $x>0$. Consider what happens as $x$ approaches the origin from the left and from the right.

You should see now why $f(-1) = f(1) = 0$ does not violate Rolle's theorem.

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Why are you writing the exponents like that? – user138246 Oct 7 '11 at 21:54
@Jordan: By definition, $a^{p/q}$ with $p$ and $q$ inters is $\sqrt[q]{a^p}$; that is, $a^{p/q} = (a^p)^{1/q}$. – Arturo Magidin Oct 7 '11 at 21:56
So $(x^2)^{-2/3}$ is equal to $2/3x^{-1/3}$? – user138246 Oct 7 '11 at 21:58
@Jordan I did write it as $(x^2)^{1/3}$ to emphasize that given a number $x$, we first square it, and them raise to power of $1/3$. This way $f(x)$ is explicitly symmetric. – Sasha Oct 7 '11 at 21:58
@Sasha: I would advice against that detour; look at Jordan's posting history, and you'll see that is far more likely to confuse him thoroughly than to make anything clear. – Arturo Magidin Oct 7 '11 at 22:04

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