The position of a particle moving along a line is given by $ 2t^3 -24t^2+90t + 7$ for $t >0$ For what values of $t$ is the speed of the particle increasing?
I tried to find the first derivative and I get
$$6t^2-48t+90 = 0$$
$$ t^2-8t+15 = 0$$
Which is giving me $ t>5$ and $0 < t < 3$, but the book gives a different answer

Let's be careful.  The velocity is $6(t^2-8t+15)$. This is $\ge 0$ when $t \ge 5$ and when $t\le 3$.  So on $(5,\infty)$, and also on $(0,3)$, this is the speed. It is not the speed on $(3,5)$.  There the speed is $-6(t^2-8t+15)$.
When $t > 5$ and also when $t< 3$, the derivative of speed is $6(2t-8)$, and is positive when $t>4$.  So the speed is certainly increasing over the interval $(5,\infty)$.
In the interval $(3,5)$, the derivative of speed is $-6(2t-8)$. This is positive in the interval $(3,4)$.
So the speed is increasing on $(5,\infty)$ and on $(3,4)$.
Note that the derivative of speed does not exist at $3$ and at $5$.
Remark: Occasionally, I have asked questions of this nature, though not quite as sadistic. Here there is a double twist. Even if the student notices that the question doesn't ask where $s(t)$ is increasing, there is the velocity/speed trap as backup. Not a good idea, it only proves one can fool most of the people most of the time.  
