Posted by **Jan** on Sunday, October 13, 2013 at 11:33pm.

Find the critical numbers of the function.

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

- Calculus - Typo -
**Steve**, Monday, October 14, 2013 at 12:02am
I think there's a typo there. Anyway, just find where h'=0.

- Calculus -
**Jan**, Monday, October 14, 2013 at 1:23am
oh, sorry about that!

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

- Calculus -
**Steve**, Monday, October 14, 2013 at 4:12am
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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