Critical Values of a Function I need to find the critical values of $h(t) = t^{3/4} - 2t^{1/4}.$
So I began by finding the derivative of the function and simplifying:
\begin{align*}
h'(t) &= (3/4)t^{-1/4} - (2/4)t^{-3/4} \\
&= \frac{3}{4t^{1/4}} - \frac{2}{4t^{3/4}}\\
&= \frac{3t^{3/4} - 2t^{1/4}}{4t}
\end{align*}
To find the first critical value, I set the denominator equal to 0 and got the first value of 0, but I'm stuck on the last critical value. The back of the book says the critical values are 0 and 4/9.
How can I go from here to find the 4/9 critical value?
Thanks for your time :)

Consider $$3t^{3\over 4}-2t^{1\over 4}=0.$$ Factoring out $t^{1\over 4}$ we obtain $$t^{1\over 4}(3t^{1\over 2}-2)=0.$$ We know that $t=0$ is a soltuion. So $$3t^{1\over 2}=2.$$ Squaring both sides we get $$9t=4.$$ Thus $$t={4\over 9}.$$ 
