Check the graph. Is f'(x)=0 anywhere on the interval?
Clearly not, so now you have to figure out why.
Is f(-1) = f(1)?
Is df/dx defined everywhere on the interval? (check out x=0)
Rolle's theorem cannot be applied to the function f(x) = x1/3 on the interval [–1, 1] because
Answer Choices:
f is not differentiable on the interval [–1, 1]
f(–1) ≠ f(1)
f is not differentiable on the interval [–1, 1] and f(–1) ≠ f(1)
Rolle's theorem can be applied to f(x) = x1/3 on the interval [–1, 1]
1 answer
1 answer