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 - 1f(x)=x3+x1?

2 Answers
Nov 8, 2016

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.

Nov 8, 2016

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)