Find the critical numbers of the function.
h(t)=t^3/4-6^1/4
3 answers
I think there's a typo there. Anyway, just find where h'=0.
oh, sorry about that!
h(t)=t^3/4-6t^1/4
h(t)=t^3/4-6t^1/4
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
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