I need help finding the critical values of this function. So $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$. So I need to set $h'(t)=0$. So for $h'(t)$ the fattest I've gotten to simplifying os $h'(t)=\frac{3}{4 \sqrt[4]{t}}-\frac{7}{4\sqrt[4]{t^3}}$ and that is as farthest as I can simplify. So i'm having a had time having $h'=0$ So could someone show me how to properly simplify and find the critical values step by step I would immensely appreciate it. In the most recent material we have covered in class ir seems that my biggest struggle comes from simplifying completely. Thanks in advance for the help. 

$h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$
$\Rightarrow$ $0=h'(t)=\frac{3}{4}t^{-\frac{1}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}+\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}}t^{\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=t^{-\frac{3}{4}}(\frac{3}{4}t^{\frac{2}{4}}-\frac{7}{4})$
Thus $t=(\frac{7}{3})^2$ Since one factor have to be $0$ and $t^{-\frac{3}{4}}$ is impossible.
