## Calculus 8th Edition

The two critical numbers are $t=0$ and $t=\frac{4}{9}$
Find the critical numbers of the function $h(t) = t^{3/4} - 2t^{1/4}$ Differentiate and set the derivative $=0$ $h'(t)=\frac{3}{4}t^{-1/4} - \frac{1}{2}t^{-3/4}$ $0=t^{-3/4}(\frac{3}{4}t^{1/2}-\frac{1}{2})$ Obviously, $h'(t) = 0$ when $t=0$, so we have our first critical number To find the other one, set the expression in the parentheses $=0$ $\frac{3}{4}t^{1/2}-\frac{1}{2}=0$ $\frac{3}{4}t^{1/2}=\frac{1}{2}$ $t^{1/2}=\frac{2}{3}$ $t=\frac{4}{9}$ This is our other critical number. Thus, the two critical numbers are $t=0$ and $t=\frac{4}{9}$