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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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    $\mathbb{R}$ is the set of all real numbers. $f(1)$ and $f(-1)$ are not functions, they are *numbers*: they are the **result** of evaluating a function, and as such they are numbers. Neither $f(1)$ nor $f(-1)$ are "functions". Just like $g(x) = x^2$ is a function, but $g(2)$ is not a function, it's a number (namely, $4$).2011-10-07

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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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    @Sasha sorry I am just not sure what I am suppose to be seeing. I am going to attempt to graph this.2011-10-08