help with differential calculus question?

So the question is asking about when john is earning points at the slowest rate possible. If P(t) represents the number of points he gets with regards to time, P'(t)= the rate that john is achieving points.

P'(t)= 4t^3-12t^2+32t+4

P'(t)=4(t^3-3t^2+8t+1)

Now you need to find when P'(t) is at it's lowest value. We see P(t) is concave up and a parabola so we know that any critical points is going to be a minimum.

Set P'(t)=0 and solve

P = t^4-4t^3-18t^2+4t+18

P ' = 4t^3 - 12t^2 - 36t + 4 << derivative .. set = 0

4t^3 - 12t^2 - 36t + 4 = 0 << divide by 4

t^3 - 3t^2 - 9t + 1 = 0 << factor

1st attempt: (t^3 - 1) - 3t(t + 9 ) << not working

2nd attempt: (t^3 - 9t) - (3t^2 -1) <<< not working

3rd : (t^3 - 3t^2) - (9t - 1) <<< no ... I do not think it factors under normal methods using integers ...

you should graph the derivative to find its zeros ... where it crosses the x axis .. then use those to find the min. value.

p=t^4-4t^3-18t^2+4t+18

dp/dt = 4t^3-12t^2-36t+4

we want the t when dp/dt is minimum

d^2p/dt^2 = 12t^2-24t-36 = 0

12(t^2-2t-3) = 0

12(t-3)(t+1) = 0

t = 3

d^3p/dt^3 = 24t -24

when t=3 , d^3p/dt^3 > 0, so dp/dt is minimum when t=3 sec

when t=3, dp/dt = -104

yes