Question

How do you find the value of c guaranteed by the mean value theorem if it can be applied for `f(x)=x^(1/3)` in the interval [-5,4]?

Answer 1

Since `f'(x) = 1/(3 root(3)x^2)` is not defined at `0` (which is in `(-5,4)`), the function is not differentiable on the interval `(-5,4)`, so we cannot use the mean value theorem.

The mean value theorem cannot be applied to this function on this interval.

Interesting Note:
Although we cannot use the mean value theorem to deduce its existence, there is, in fact, a value of `c` in the interval `(-5,4)` with `f'(c)=(f(4)-f(-5))/(4-(-5))`.

`c = sqrt(3/(root(3)4+root(3)5))^3` which is about `0.8678` will work.

(Yes, I know I can rationalize the denominator of `3/(root(3)4+root(3)5)`, but that's not what this note is about.)