Answer
a. $t=3$
b. $t=4$
Work Step by Step
The original function is $ f(t)=t^3-27t+5 $ therefore, $ f'(t)=3t^2-27$
a. Set $t=0$ from the function $ f'(t)=3t^2-27$
$ 3t^2-27=0$
$ 3t^2=27$
$ t^2=9$
$ t=3$
b. Set $t=21$ from the function $ f'(t)=3t^2-27$
$ 3t^2-27=21$
$ 3t^2=48$
$ t^2=16$
$ t=4$
