Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 70

If the distance function is $s(t) = t^4 + t^3 - 4t^2 - 2t + 4$, illustrate $s, v$ and $a$ over the
interval $[-3,3]$. Then use the graph to determine the point(s) at which the velocity will switch
from increasing to decreasing or from decreasing to increasing.

We have, $s(t) = t^4 + t^3 - 4t^2 - 2t + 4$, so

$
\begin{equation}
\begin{aligned}
v(t) &= s'(t) = 4t^3 + 3t^2 - 8t - 2\\
\\
a(t) &= v'(t) = 12t^2 + 6t - 8
\end{aligned}
\end{equation}
$


Then, the graph is



Based from the graph, the velocity switches at $t \approx -1.25$ from increasing to decreasing.
On the other hand, the velocity switches at $t \approx 0.60$ from decreasing to increasing. These values
are obtained by looking at the peaks of the graph of the velocity function or by looking at the values where
the acceleration function crosses the $x$-axis because the slope and time peaks of the velocity is zero.