Calculus
posted by
Jan
.
Find the critical numbers of the function.
h(t)=t^3/46^1/4

Calculus  Typo 
Steve
I think there's a typo there. Anyway, just find where h'=0.

Calculus 
Jan
oh, sorry about that!
h(t)=t^3/46t^1/4

Calculus 
Steve
well, dh/dt = 3t^2(11∜t8) / 8(23∜t)^2
We want dh/dt=0, so t=0 or t=(8/11)^4
h(t) and dh/dt are undefined where t=(2/3)^4