correct. f'(x) = 2/3∛x
which is not defined at x=0.
so, the requirements of the theorem are not met. There is, in fact no value of x where f'(x) = 0.
Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(x)=0.
f(x) = x^(2/3) - 1 [-8,8]
I plugged in both values and found out that they both equal 4 so I thought Rolle's theorem applied.
However, the answer is that this function is not differentiable at 0.
Can someone explain to me why it is not differentiable even though f(a)=f(b).
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