# Given the function f(x)=x^3-3x+2, how do you verify that f satisfies the hypotheses of the Mean Value Theorem on [-2, 2]?

## What are all the numbers $c$ that satisfy the conclusion of the MVT?

Dec 4, 2017
1. Check that $f \left(x\right)$ is continuous & differentiable on the interval.
2. Find the values of $c$ (see below).

$c = \pm \sqrt{\frac{4}{3}} = \pm 1.1547 \ldots$

#### Explanation:

The main conditions of the Mean Value Theorem are:

• f(x) must be continuous on [a, b]
• f(x) must be differentiable on (a, b).
• There is some point $c$ on (a, b) that satisfies:

$f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$

Translating that 3rd postulate: there must be some point on this interval such that the instantaneous rate of change there is equal to the average rate of change on that interval.

For your function, you can assume that conditions 1 and 2 are fulfilled since you have a polynomial, and polynomials are differentiable (and continuous) everywhere . There is a more rigorous way of proving differentiability & continuity, but for this function there is no need to do that.

For condition 3, the first thing we need to do is find that value of $c$, and then check whether or not it is on our interval (which it should be). So we set up the following:

$\frac{d}{\mathrm{dx}} \left({x}^{3} - 3 x + 2\right) = \frac{f \left(2\right) - f \left(- 2\right)}{2 - \left(- 2\right)}$

$3 {x}^{2} - 3 = \frac{4 - 0}{4} = 1$

Now we have an equation which we can solve to find our value for $c$!

$3 {x}^{2} - 3 = 1$

$3 {x}^{2} = 4$

$x = c = \pm \sqrt{\frac{4}{3}} = \pm 1.1547 \ldots$

As it turns out, there are two values of c, and they are both on our interval!

If you'd like further clarification on the Mean Value Theorem - as well as its "spin-off" Rolle's Theorem - check out this excellent video by Professor Leonard:

Hope that helps :)