show that the mean value theorem applies and find the value of c guaranteed by the theorem

a = 0 b = 2

f(x) = x^3 - x

f '(x) = 3x^2 - 1

f(a) = (a)^3 - (a) = (0)^3 - (0) = 0

f(b) = (b)^3 - (b) = (2)^3 - (2) = 8 - 2 = 6

Now the formula for mean value is

f '(x) = f(b) - f(a) / b - a

f '(x) = 6 - 0 / 2 - 0

f '(x) = 6 / 2

f '(x) = 3

Now to find the value of c

Put x = c in f'(x) = 3x^2 - 1 , then

f'(c) = 3c^2 - 1

Now

3c^2 - 1 = 3

3c^2 = 3 + 1

3c^2 = 4

c^2 = 4/3

c = +√4/3 or c = -√4/3

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up