Asked by Jan

Find the critical numbers of the function.

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

Answers

Answered by Steve
I think there's a typo there. Anyway, just find where h'=0.
Answered by Jan
oh, sorry about that!

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