Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 10

a.) Graph the function $f(x) = x^3 - 2x$ and its secant line through the points $(-2,-4)$ and $(2,4)$ on the viewing rectangle $[-3,3]$ by $[-5,5]$ use the graph to estimate the $x$-coordinates of the points where the tangent line is parallel to the secant line
b.) Determine the exact values of the numbers $c$ that satisfy the conclusion of the Mean Value Theorem for the interval $[-2,2]$ and compare your answer in part(a).

a.)



Based from the graph, the $x$-coordinate where the tangent line is parallel to the secant line can be approximated as $x \approx 1.15$ and $x \approx -1.15$

b.) Solving for the exact value of $c$, we get...

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


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

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

Hence, the exact values of $c$ are $\displaystyle c = \sqrt{\frac{4}{3}}$ and $\displaystyle c = - \sqrt{\frac{4}{3}}$ which is close to our guess in part(a).