Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 52

Determine at which points on the curve $y = 1 + 40 x^3 - 3x^5$ does the tangent line have the longest slope?
Taking the derivative of $y$, we get...
$y' = 120x^2 - 15x^4$

We take the derivative of $y'$, so...
$y'' = 240x - 60x^3$

when $y'' = 0$

$
\begin{equation}
\begin{aligned}
0 &= 240x - 60x^3\\
\\
0 &= x(240-60x^2)
\end{aligned}
\end{equation}
$

We have,
$x = 0 \text{ and } 240 - 60x^2 = 0\\
x= 0 \text{ and } x = 2, \, x= -2$


If we evaluate $y'$ with the $x = 0$, $x = 2$, and $x = -2$, then


$
\begin{equation}
\begin{aligned}
y'(0) &= 120(0)^2 - 15(0)^4\\
\\
y'(0) &= 0\\
\\
y'(2) &= 120(2)^2 - 15(2)^4\\
\\
y'(2) &= 240\\
\\
y'(-2) &= 120(-2)^2 - 15(-2)^4\\
\\
y'(-2) &= 240
\end{aligned}
\end{equation}
$

Therefore, we can say that the tangent line have the largest slope when $x = 2$ or $x = -2$.