Question

Consider the function `f(x) = 4 x^3 − 4 x` on the interval [ −3 , 3 ], how do you find the average or mean slope of the function on this interval?

Answer 1

See the explanation below

Explanation:

The average or mean slope of a function `f(x)` on the interval `[a,b]` is

`=(f(b)-f(a))/(b-a)`

Here,

`f(x)=4x^3-4x=4x(x^2-1)`

and the interval is `=[-3,3]`

`f(3)=12(8)=96`

`f(-3)=-12*8=-96`

Therefore,

The mean slope is

`=(f(3)-f(3))/(3-(-3))=(96+96)/(6)=32`

Then,

There exists `c in (-3,3)` such that

`f'(c)=32`

`f'(x)=12x^2-4=4(3x^2-1)`

Therefore,

`f'(c)=4(3c^2-1)=32`

`3c^2-1=8`

`c^2=9/3=3`

`c=+-sqrt3`

So,

`+-sqrt3 in (-3,3)`