Question

Given the function `f(x) = x^(1/3)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-5,4] and find the c?

Answer 1

See below.

Explanation:

The Mean Value Theorem has two hypotheses:

H1 : `f` is continuous on the closed interval `[a,b]`

H2 : `f` is differentiable on the open interval `(a,b)`.

In this question, `f(x) = x^(1/3) = root(3) x` , `a= -5` and `b= 4`.

This function is a power function, so it is continuous on its domain, `(-oo, oo)`, so it is continuous on `[-5, 4]`. The function does satisfy the first hypothesis on this interval

`f'(x)=1/3x^(-2/3) = 1/(3(root(3)x)^2)` which exists for all `x != 0` and which does not exist for `x = 0`. Since `0` is in `(-5,4)` the function does not satisfy the second hypothesis on this interval.
(That is, there is at least one point in `(-5,4)` at which `f` is not differentiable.)

Although this function does not satisfy the hypotheses on the interval, it does satisfy the conclusion (in two places).

Solving

`f'(x) = (f(4)-f(-5))/(4-(-5))` leads to

`1/(3(root(3)x)^2) = (root(3)4 - root(3)(-5))/9`

And after some tedious algebra, we get

`x = +- sqrt((-11+4(10^(1/3)) +10^(2/3))/3)`

`~~ +- 0.87`

both of which are in `(-5,4)`.

Although this is tedious to work out analytically, it look likely geometrically.

Here is a graph of the function and the secant line

graph{(y-x^(1/3))((y-4^(1/3))/(x-4) - (4^(1/3)+5^(1/3))/9) = 0 [-6.79, 5.7, -4.444, 1.8]}

It certainly looks as if there ought to be two places where the tangent is parallel to the secant. And there are.