Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 12

Verify that the function $f(x) = x^3 + x - 1$ satisfies the hypothesis of the Mean Value Theorem on the interval $[0,2]$. Then find all the numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
We know that $f(x)$ is a polynomial function that is continuous everywhere and its derivative, $f'(x) = 3x^2 + 1$ is a quadratic function that is continuous on every values of $x$ as well. Hence, $f(x)$ is continuous on the closed interval $[0,2]$ and differentiable on the open interval $(0,2)$.

Now solving for $c$, we have...

$
\begin{equation}
\begin{aligned}
f'(c) &= \frac{f(b) - f(a)}{b - a }\\
\\
f'(c) &= \frac{\left[ (2)^3 + 2 - 1\right] - \left[ (0)^3 + 0 - 1\right]}{2- 0}\\
\\
f'(c) &= 5
\end{aligned}
\end{equation}
$

but $f'(x) = 3x^2 + 1$, so $f'(c) = 3(c)^2 +1$

$
\begin{equation}
\begin{aligned}
3c ^2 + 1 &= 5\\
\\
3c^2 &= 4\\
\\
c^2 &= \frac{4}{3}\\
\\
c &= \pm \sqrt{\frac{4}{3}}
\end{aligned}
\end{equation}
$

We got two values of $c$ but the function is defined only on interval $[0,2]$. Hence, $\displaystyle c = - \sqrt{\frac{4}{3}}$ is disregarded. Therefore, $\displaystyle c = \sqrt{\frac{4}{3}}$