Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem f(x)=x^3-2x^2f(x)=x32x2; [0, 2]?

1 Answer
Oct 13, 2016

The conclusion of the Mean Value Theorem for ff on [a,b][a,b] says:

there is a cc in (a,b)(a,b) such that f'(c) = (f(b)-f(a))/(b-a)

(The theorem make no guarantees about our ability to find the value(s) of c)

Find the value(s) of c does not use the Mean Value Theorem, it uses the derivative and some algebra.

For f(x) = x^3-2x^2 on [0,2], the conclusion of MVT says:

there is a c in (0,2) such that f'(c) = (f(2)-f(0))/(2-0)

To find the c (or c's) find f'(x), do the arithmetic on the right and solve the resulting equation. In this case, solve

3x^2-4x = 0

The solutions are 0 and 4/3.

The c mentioned in MVT must be in (0,2), so 4/3 is the only value of c