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)#?

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