#### Answer

The two critical numbers are $t=0$ and $t=\frac{4}{9}$

#### Work Step by Step

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 $t=\frac{4}{9}$