Calculus position/velocity question. Disclaimer: I am not a student trying to get free internet homework help.  I am an adult who is learning Calculus.  I am deeply grateful to the members of this community for their time.
Here is the question (Not sure of the best way to format this)
$$s(t)=t^3-9t^2+24t-6$$
Q:  The speed of the particle is decreasing for:
1) t<1
2) t>2
3) t<3
4) t<1 and t>2
5) all t
So, I took the derivative of position to get velocity, and set it to zero.
$$v(t)=3t^2-18t+24=0$$
Now, I have a parabola with roots at (2,0) and (4,0) and vertex (3,-3)
I understand that Speed is DECREASING when either:
a) velocity is (+) and slope is (-)
(I'm moving forward, but velocity is becoming less positive)
b) velocity is (-) and slope is (+)
(I'm moving backwards, but velocity is becoming less negative)
So, MY answer is not in the list $t<2$ and $3<t<4$
$t<2$ is where the parabola meets criterion (a)
$3<t<4$  is where the parabola meets criterion (b)
Yet, the answer key says choice (3) ??
What am I doing wrong?
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Update: 
There was another question in the same set that is equally perplexing!
Q:  The minimum value of the speed is:
1) -3
2) -2
3) -1
4) 0
5) 1
MY answer would be 0.  (choice 4), since speed = |v|
Yet, the answer key says choice 3
How could speed OR velocity be equal to -1 ?
Minimum velocity is -3 (choice 1)
 A: It appears the answer key is using velocity instead of speed.  Following your definition of speed as absolute value of velocity, you are correct that it is decreasing on $t \lt 2$, but it is also decreasing on $3 \lt t \lt 4$ (your case b).  That is still not one of the choices.  It looks like you understand the issue, though.
A: Differentiate one more time t0 get the acceleration, $a(t) = 6t-18$. The speed is decreasing when $a(t) <0$, or in other words, when $t < 3$.

A: I think what's going on here is most likely this:
If the position is given by
$s(t)=t^3-9t^2+24t-6, \tag{1}$
then as you correctly found the velocity is the derivative of $s$:
$v(t)=3t^2-18t+24. \tag{2}$
$v(t)$ is decreasing when $v'(t) < 0$; we have
$v'(t) = 6t - 18, \tag{3}$
so 
$v'(t) < 0 \Leftrightarrow t < 3, \tag{4}$
in agreement with your answer key. 
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: The question is whether the speed is decreasing. Speed is a scalar, while velocity is a vector. You need to see where the speed is moving away from zero
$s(t) = t^3 - 9t^2 + 24t - 6$
$v(t) = s'(t) = 3t^2 - 18t + 24$
Set velocity equal to 0 and solve
$0 = 3t^2 - 18t + 24$
$0 = t^2 - 6t + 8$
$0 = (t-4)(t-2)$
From here we can see that the vertex is located at $x=\dfrac{2+4}{2}=3$. The parabola opens upwards because the leading coefficient is positive. However, speed is always positive, and is the absolute value of velocity:
$speed(t) = |v(t)|$
What can we deduce from this?


*

*The speed is positive and decreasing on $(-\infty,2)$

*The speed is 0 at $x=2$

*The speed is positive and increasing on $(2,3)$

*The speed is constant at $x=3$

*The speed is positive and decreasing on $(3,4)$

*The speed is 0 at $x=4$

*The speed is positive and increasing on $(4,\infty)$


Why is the speed always positive? Because speed is a scalar. Speed is the absolute value of velocity. The speed is therefore decreasing on $(-\infty,2)\cup(3,4)$. The question is wrong to ask about speed.
Your question should be asking you when velocity is decreasing. An inspection of the graph will show us that it decreases from $-\infty$ to 3 before it starts going back up again.
But here's the calculus way to get that answer - take the second derivative:
$a(t) = v'(t) = 6t-18$
The acceleration shows us the change in velocity. It is negative from $-\infty$ until its zero at $t=3$, which means that velocity is decreasing. That is, the change in velocity is negative.
