Question

Find the exact values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval [−2, 2]. (Enter your answers as a comma-separated list.)

Answer 1

Answer:

The answer is "$\bold{c= \pm \frac{2}{\sqrt{3}}}$"

Step-by-step explanation:

If the function is:

$\to f'(x) = 3x^2-2 \\\\\to f'(c) = 3c^2-2$

points are:

$\to -2 \leq x \leq2$

use the mean value theorem:

$\to f'(c) = \frac{ f(b)- f(a)}{b-a}$

            $= \frac{ f(2)- f(-2)}{2-(-2)}\\\\= \frac{4-(-4) }{4}\\\\= \frac{8}{4}\\\\= 2$

$\to 3c^2-2=2 \\\\\to 3c^2=4 \\\\\to c^2=\frac{4}{3} \\\\c= \pm \frac{2}{\sqrt{3}}$