Answer
$x=0$ and $x=4$
Work Step by Step
An interior point of the domain is a critical point of $f$ if
$f'$ is zero or undefined.
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$y=x^{2}-32x^{1/2}$
$y'=2x-32\displaystyle \cdot\frac{1}{2}x^{-1/2}=2x-\frac{16}{\sqrt{x}}$
$x=0$ is a critical point as $y'$ is undefined there
Checking where the derivative is zero,
$2x-\displaystyle \frac{16}{\sqrt{x}}=0$
$\displaystyle \frac{2x\sqrt{x}-16}{\sqrt{x}}=0\quad\Rightarrow\quad 2x\sqrt{x}-16=0$
$\Rightarrow\quad 2x^{3/2}=16$
$\Rightarrow\quad x^{3/2}=8$
$\Rightarrow\quad x=8^{2/3}=[(2^{3})^{1/3}]^{2}=2^{2}=4$
$x=4$ is a critical point
