Question

How do you determine all values of c that satisfy the conclusion of the mean value theorem on the interval [0, 8] for `f(x) = x^3 + x - 1`?

Answer 1

Solve the equation `f'(x) = (f(8)-f(0))/(8-0)`. Every solution in `(0,8)` is a permissible value of `c`.

Explanation:

`f'(x) = 3x^2+1`.

`(f(8)-f(0))/(8-0) = 65`.

`3x^2+1 = 65` if and only if `x= +- 8/sqrt3`. The negative solution is not in the interval.

The only value of `c` that satisfies the conclusion is `8/sqrt3`.

Answer 2

x= `+-8/sqrt(3)`

Explanation:

The mean value theorem states: `(f(b)-f(a))/(b-a)`

so you have `(f(8)-f(0))/(8-0)`

Plug 8 and 0 into the equation `f(x)= x^3 + x -1`

f(8)= 512 + 8 - 1 = 519
f(0)= 0 + 0 - 1 = -1

`(519+1)/(8-0)` = 65

Now take the derivative of the function `f(x)= x^3 + x -1`

`3x^2+1` is the derivative, now set it equal to 65.
`3x^2+1`= 65, subtract the 1 to the other side
`3x^2`= 64, divide the 3 to the other side
`x^2`= `64/3`, take square of both side

x= `sqrt(64)/sqrt(3)`

x= `+-8/sqrt(3)`