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 :)

You do not set the denominator to zero, that would give you a division by zero, which is undefined, rather than zero.
You always set the numerator equal to zero. Then note that $t=0$ will make the numerator zero, or, if $t\neq0$, then you can divide by $t$ and find $t=4/9$.
The $t=0$ case is called a trivial solution and is something to watch out for when solving any type of equation.
