Question

Given the function `f(x)=(x-3)^(2/3)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and find the c?

Answer 1

Please 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)`.

We say that we can apply the Mean Value Theorem if both hypotheses are true.

H1 : The function `f` in this problem is continuous on `[1,4]`
Because power functions are continuous on their domains and linear functions are continuous everywhere. And the composition of continuous functions is continuous.

H2 : The function `f` in this problem is not differentiable on the entire interval `(1,4)`
Because the derivative, `f'(x) = 2/3(x-3)^(-1/3)` fails to exist at `3` which is in the interval `(1,4)`.

A note on "if . . . then . . . " theorems

If the hypotheses are not true, we learn nothing about the truth of the conclusion.

To determine whether the conclusion is true or false we try to solve the equation in the conclusion of MVT.

Solving `f'(x) = (f(4)-f(1))/(4-1)` yields one solution of about `-39.47` which is not inside the interval `(1,4)`.

Here is a graph of the function and the secant line through the points with `x=1` and `x=4`. You can scroll in and out and drag the viewing window with a mouse. When you leave the page and return, the graph will have reset to the default shown the first time.

graph{(y-(x-3)^(2/3))((y-1)-(1-4^(1/3))/(3)(x-4))=0 [-15.1, 10.21, -5.03, 7.63]}