Question

Verify that the function ="g(x)=2x^3-3x+1" align="absmiddle" class="latex-formula"> satisfies the three hypotheses of Rolle’s Theorem on the interval $[0,2]$.

Answer 1

Lets check if the three conditions hold.

1 : Continuity of g on the interval [0,2]

First, g(x) is a continuous function on R, as the sum of a cubic function wich is continuous on R, and a linear polynomial of the form ax + b which is also continuous on R. Finally g is also continuous on the interval [0,2]

2 : Differentiable on the same interval

Since the cubic function and the linear polynomial one are differentiable on R, g also is differentiable and particularly on the interval [0,2]

Also we have g'(x) = 2*3*x² - 3 = 6x² - 3

3 : Do we have g(0) = g(2) ?

Lets compute g(0) = 2*0^3 - 3*0 + 1 = 1

And g(2) = 2*2^3 - 3*2 + 1 = 2 * 8 - 6 + 1 = 16 - 6 + 1 = 11

Since g(0) ≠ g(2), Rolle's theorem is not applicable. Thus unfortunately, we can not conclude that there exist c ∈ (0,2) such that f'(c) = 0