Question

critical points for f(t)=8t^3-t^2

Answer 1

To find critical points you take the derivative and set it equal to zero:
f'(t) = 24t^2-2t
0 =t(24t - 2)
t = 0 or 1/12

Answer 2

\[ f(t)=8t^3-t^2\]\[ f'(t)=24t^2-2t=t(24t-2)=0\]\[t=\left\{ 0,1/12 \right\}\] continuing for the second derivative...

Answer 3

\[f''(t)=48t-2=0\]\[t=\left\{ 1/24 \right\}\]