Rolle's theorem violation or not?

Question

Here is a question that I am trying to solve. Unsure about the best way to go about it. Could use some help:

Question Let $f(x) = 1 - x^{\frac{2}{3}}.$ Show that $f(-1) = f(1)$ but there is no number c in (-1, 1) such that $f'(c) = 0$. Why does this not contradict Rolle’s Theorem?

Answer explanation

So Rolle's thereom says that if f(x) is continuous and differentiable and f(a) = f(b), then there is a number c in between a and c where the slope of f(x) is = 0.

First off, $f(-1) = 1 - 1 = 0$ and $f(1) = 1 - 1 = 0$

Side question Quick question, why is that when I plug in $(-1)^{2/3}$ into google it does not give me -1. Am I wrong in saying that $(-1)^{2/3} = 1$?

So since f(1) and f(-1) both = 0, then I'm going to try to see if maybe the slope at f'(0) is = 0.

So here are a couple of ways that I determined this. Are both valid?

Method 1

$$f'(x) = -\frac{2}{3} x^{-1/3}$$

And that is undefined at x = 0 right? So since f(x) is not differentiable at x = 0, then this function isn't differentiable, and so Rolle's theorum is not violated.

Method 2

$\lim_{x \to 0^{+}} f'(x)$ is negative

$\lim_{x \to 0^{-}} f'(x)$ is positive

Does this show that the differential at 0 does not exist?

Answer 1

None of the two methods really proves that $f$ is not differentiable at $0$.

Method $1$ only shows that the formula for the derivative is not valid at $0$, and method $2$ only shows that $f'$ is not continuous at $0$.

Both methods are still useful, though. It is clear that if $f$ is not differentiable at some point in $[-1,1]$, this point must be $0$.

You should prove that

$$\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{h^{2/3}}h$$

does not exist.

Answer 2

You're right that the function is not differentiable at $0$, and that is why Rolle's theorem does not apply. Your method 1 is not quite sufficient to show it, though: It could conceivably be that the function is differentiable at $0$, even though the expression you get from differentiating is not defined there. (Example: $f(x)=x^2\sin(1/x)$.

Your method 2 is better, but it needs more work too: To make use of the different limits of the derivatives from each side, you need to use L'Hôpital's rule to show that the two one-sided derivatives equal those limits. And since they are different (they are in fact $\pm\infty$), the function is not differentiable there. (This requires continuity, but you don't need to prove continuity, for if the function weren't continuous, it definitely wouldn't be differentiable.)

Finally, the most basic method is to employ the definition of derivative directly: $$ f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^{2/3}}{x}=\lim_{x\to0}x^{-1/3}$$ does not exist.