Verify that the function f(x)=3x^2 + 2x + 5 satifies the hypotheses of the MVT on the interval[-1, 1]?

Question

Verify that the function f(x)=3x^2 + 2x + 5 satifies the hypotheses of the Mean Value Theorem on the interval [-1, 1]. then find all number(s) C that satisfy the conclusion of the Mean Value theorem.

Answer 1

The mean value states that for a function differentiable on the closed interval [a, b], ∃c∈(a, b) such that f'(c)=(f(b)-f(a))/(b-a). So first compute (f(b)-f(a))/(b-a), and then find all numbers c in (a, b) such that f'(c) is equal to it.

f(1) = 10
f(-1) = 6
1-(-1) = 2
(f(1)-f(-1))/(1-(-1)) = (10-6)/2 = 2

So we are looking for all numbers x in (-1, 1) such that f'(x) = 2.

f'(x) = 6x+2 = 2 → 6x = 0 → x=0

So the sole number satisfying the MVT on [-1, 1] is 0