Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 2

Show that the function $f(x) = x^3 - x^2 - 6x + 2$ satisfies the three hypothesis of Rolle's Theorem on the interval $[0,3]$. Then find all numbers $c$ that satisfy the conclusion of Rolle's Theorem.

We know that $f(x)$ is polynomial function that is continuous everywhere. Hence, $f(x)$ is continuous or the closed interval $[0,3]$
Next, if we take the derivative of $f(x)$, we get...
$f'(x) = 3x^2 - 2x - 6$

We also know that $f'(x)$ is a quadratic function that is differentable everywhere, hence, $f$ is differentiable on the open interval $(0,3)$
Lastly, if $f(0) = f(3)$
$(0)^3 - (0)^2 - 6(0) + 2 = (3)^3 - (3)^2 - 6(3) + 2$
$2 = 2$
Since we satisfy all the hypothesis of Rolle's Theorem, we can now solve for $c$ where $f'(c) = 0$, so..
$f'(c) = 3x^2 - 2x - 6 = 0$
$0 = 3x^2 - 2x - 6 = 0$


By using Quadratic formula, we get...
$\displaystyle x = \frac{1+\sqrt{19}}{2}$ and $\displaystyle x = \frac{1 - \sqrt{1}}{3}$
$ x = 1.7863$ and $ x = -1.1196$

The function is defined only at interval $[0,3]$, therefore, the value of $c$ is $c = 1.7863$