Consider the function f(x)=4−4x^(2/3) on the interval [−1,1]. Which of the three hypotheses of Rolle's Theorem

Consider the function f(x)=4−4x^(2/3) on the interval [−1,1].

Which of the three hypotheses of Rolle's Theorem fails for this function on the inverval?

(a) f(x) is continuous on [−1,1].
(b) f(x) is differentiable on (−1,1).
(c) f(−1)=f(1).

The answer is B.
But I was wondering if C and A were wrong, how would you go about finding why they fail for the function?

The way you can determine if parts a) and b) of Rolle's Theorem are false is in part by knowing what types of functions INFALLIBLY are continuous and differentiable. All polynomial and rational functions are. A lot of trigonometric functions (their inverses, no).

Checking for the holding up of the conditions for Rolle's Theorem, there are 3 conditions:
- the function is constant
- function is greater than f(a) for some number in (a,b)
- function is less-than f(a) for some number in (a,b)

I've seen this used to argue the number of roots in what is called "proof by contradiction" where you take the derivative of the function, and solve for "x": if it can't be done, then that's justification for stating there are only "n" root(s).

This don't pop up again in the near future in calculus, so don't sweat this too much. A lot of these dime explanations are a formality, that good textbooks and courses supply strong geometric arguments to replace. Basically, just note that unless its constant, a function will reach a maximum/minimum value where there is a number "n" that has f '(n) = 0 (meaning slope of zero, also meaning maximum/minimum).