Question

How do you determine all values of c that satisfy the mean value theorem on the interval [-1,1] for `(x^2 − 9)(x^2 + 1)`?

Answer 1

`c=0`

Explanation:

First, we need to figure out what the points are, so we know what the slope between them is.

`f(-1) = ((-1)^2-9)((-1)^2+1) = (1-9)(1+1) = -16`

`f(1) = ((1)^2 - 9)((1)^2 + 1) = (1-9)(1+1) = -16`

The two points have the same value, so the slope between them is zero.

The mean value theorem says that:

If the slope between two points on a graph is `m`, then there must be some point `c` between those points at which the derivative is also `m`.

In this case, our slope is 0, so we're looking for points between `-1` and `1` whose slope(s) is/are 0.

To do this, let's take the derivative of our function.

`f'(x) = d/dx(x^2-9)(x^2+1)`

`= 2x(x^2+1) + 2x(x^2-9)`

`= 2x^3 + 2x + 2x^3 - 18x`

`= 4x^3 - 16x`

This derivative represents our slope, so we can set it equal to 0, since we're looking for where the slope is 0.

`4x^3 - 16x = 0`

`x^3 - 4x = 0`

`x(x^2-4) = 0`

`x(x-2)(x+2) = 0`

`therefore x in {-2, 0, 2}`

The three points where the slope is zero are `-2`, `0`, and `2`. However, since our problem wants us to find points we can use for the MVT for `-1` and `1`, we can only choose points between `-1` and `1`. Therefore, the only point we can use is 0.

`c=0`

Final Answer