Number of times particle reverses direction on $s = t^5 + 5t^4$ The question I'm solving is "If a particle moves along a line according to the law $s = t^5 + 5t^4$, then the number of times it reverses direction is:" with the answers $0, 1, 2, 3, 4$. I can see from the graph it reverses twice, but I'd have to solve this without a calculator for my exams.
Now according to Barron's, "If $s$ is a continuous function of $t$, the particle reverses direction whenever $v$ (velocity) is zero and a (acceleration) is NOT zero". Using this rule, I computed the velocity function as $v = 5t^4 + 20t^3$ and acceleration as $a = 20t^3 + 60t^2$. 
Velocity is zero at $t = -4$, $t = 0$; however, acceleration is also $0$ at $t = 0$, yet the graph still reverses. So how would I go about solving this without graphing the original equation? 
 A: The statement you reference, namely

If $s$ is a continuous function of $t$, the particle reverses direction whenever $v$ (velocity) is zero and a (acceleration) is NOT zero.

is true, however, the opposite direction is not true, because it could be that it reverses direction even if velocity and acceleration are both zero. You've already seen that this problem is an example of this.
The easiest way of dealing with this issue (in most cases) is simply not to compute acceleration at all: instead, look at where the derivative changes sign. (Sometimes this is done by drawing a "sign line".)
The particle reverses direction if and only if the velocity changes sign.
In this case, we have
$$
v = 5t^3(t+4)
$$
You've observed that this is zero at $t = 0$ and $t = -4$. But does it switch sign at both these points? To find out, simply plug in some values in between and on either side
\begin{align*}
\text{At } &t = -5, v = 5(-5)^3(-5 + 4) > 0. \\
\text{At } &t = -2, v = 5(-2)^3(-2 + 4) < 0. \\
\text{At } &t = 1, v = 5(1)^3(1 + 4) > 0. \\
\end{align*}
So we have that velocity is positive for $t < -4$, negative for $-4 < t < 0$, and positive for $t > 0$. That means it switches sign at both $-4$ and $0$, so both of those are points where the particle reverses direction.

Edit: There's another method that works, too, called the higher-order derivative test. In this case, the fourth derivative of the particle's position, $120 + 120t$ is positive at $t=0$, and four is even, which means the velocity changes from negative to positive. You can read about this test here.
