How do you determine whether the function satisfies the hypotheses of the Mean Value Theorem for #f(x)=x^(1/3)# on the interval [-5,4]?

There are two hypotheses for MVT.

The function must be continuous on #[-5,4]#?

The function must be differentiable on #(-5,4)#?

The first is true (satisfied) because #x^(1/3)# is the 3rd root function and a root function is continuous on its domain. In this case the domain is #RR#, so it includes #[-5,4]#.

This function fails to be differentiable at #0#, so the second hypothesis is not satisfied on #(-5,4)#.

This function does not satisfy the hypotheses of the Mean Value Theorem on this interval.

Bonus material

This function DOES satisfy the conclusion of the MVT on this interval. We cannot use the Mean Value Theorem to conclude that there is a #c# in #(-5,4)# such that #f'(c) = (f(4)-f(-5))/(4-(-5))#.

We can, however solve #f'(c) = (f(4)-f(-5))/(4-(-5))# algebraically. We find that there are two solutions in the interval.

The algebra is be tedious to type, but the graph makes this plausible.
It shows #f(x) = x^(1/3)# and the secant line joining the points with #x=-5# and #x=4#.

graph{(y-x^(1/3)) (y-4^{1/3)-((4^(1/3)+5^(1/3))/9)(x-4))= 0 [-6.06, 5.04, -2.864, 2.685]}