Question

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.) f(x) = x3 − x2 − 6x + 2, [0, 3] c =

Answer 1

Answer:  $\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}$

Step-by-step explanation:

There are 3 conditions that must be satisfied:

1. f(x) is continuous on the given interval
2. f(x) is differentiable
3. f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2     [0, 3]

1. There are no restrictions on x so the function is continuous $\checkmark$

2. f'(x) = 3x² - 2x - 6 so the function is differentiable $\checkmark$

3. f(0) =  0³ - 0² - 6(0) + 2 = 2

   f(3) =  3³ - 3² - 6(3) + 2  = 2

   f(0) = f(3) $\checkmark$

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1$

Only one of these values (1.8) is between 0 and 3.