Question

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (Select all that apply.)
f(x) = sqrt((2-x**(2/3))**3) text(, [-1,1])

Answer 1

$f(x)$ has to be differentiable on [-1, 1] for the theorem to hold. This is not the case because the derivative doesn't exist at $x=0$.

$f(x)=\sqrt{\left(2-x^{2/3}\right)^3}=\left(2-x^{2/3}\right)^{3/2}$
$\implies f'(x)=\dfrac32\left(2-x^{2/3}\right)^{1/2}\left(-\dfrac23x^{-1/3}\right)=-\dfrac{\left(2-x^{2/3}\right)^{1/2}}{x^{1/3}}$

but clearly $f'(0)$ is undefined.