Question

Finding roots of a function with mean value theorem?

I am suppose to show that the equation `x^3−15x+c=0` has at most one root in the interval `[-2,2]`

I have sort of memorized the mean value theorem but I don't really understand how it is applicable to this.

Answer 1

Please see below. (Also see the second part of Example 2 in Stewart's Calculus section 3.2 if that's the book you're using.)

Explanation:

The Mean Value Theorem (MVT) and similar Rolle's Theorem are not used to find zeros of functions.

But they can be used to prove that statement requested.

Given a constant `c`, suppose that `x^3-15x+c=0` has two different solutions and suppose that they are both in the interval `[-2,2]`.
Call the solutions `x_1` and `x_2` with `x_1 < x_2`

Let `f(x) = x^3-15x+c` and note that `f(x_1) = f(x_2) = 0`.

`f` is a polynomial so it is continuous and differentiable everywhere.
In particular,
(1) `f` is continuous on `[x_1,x_2]` and
(2) `f` is differentiable on `(x_1,x_2)`.

Therefore, by MVT there is a number `k` in `(x_1,x_2)` with

`f'(k)=(f(x_2)-f(x_1))/(x_2-x_1) = 0` `" "` (See Note 1 below.)

But that means there is a `k` in `[-2,2]` with

`f'(k) = 3k^2-15=0`. `" "` (See Note 2 below.)

However, the only solutions to `3x^2-15=0` are `+-sqrt5`, neither of which is in the interval `(-2,2)`.

Therefore, there cannot be two different solutions to `x^3-15x+c=0` in the interval `[-2,2]`.

(That is: there is at most one solution o `x^3-15x+c=0` in the interval `[-2,2]`.)

Note 1:

Instead of using MVT, we could use:

(3) `f(x_1)=f(x_2)` by hypothesis.

So, by Rolle's Thereom, there is a number `k` in `(x_1,x_2)` with

`f'(k) = 0`

Note 2:

We started with `k` in `(x_1,x_2)`, but `(x_1,x_2)` is inside `(-2,2)`,

so `k` is also in the larger interval, `(-2,2)`.