Question

Find the critical numbers of the function.

h(t)=t^3/4-6^1/4

Answer 1

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

Answer 2

oh, sorry about that!

h(t)=t^3/4-6t^1/4

Answer 3

well, dh/dt = 3t^2(11∜t-8) / 8(2-3∜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