# What are the values and types of the critical points, if any, of f(x) =x^2-sqrtx?

May 11, 2018

$x = 2 \sqrt[3]{2}$ is the minimum.

#### Explanation:

f(x) =x²-sqrtx
We search $f ' \left(x\right) = 0$
$f ' \left(x\right) = 2 x + \frac{1}{2 \sqrt{x}}$
$2 x + \frac{1}{2 \sqrt{x}} = 0$
$2 x \sqrt{x} = - \frac{1}{2}$
$x \sqrt{x} = - \frac{1}{4}$
x³=1/16
$x = 2 \sqrt[3]{2}$
Let $X = \sqrt{x}$, we can see f(X)=X⁴-X, and so that $x = 2 \sqrt[3]{2}$ as a minimum of $f \left(x\right)$ (because f take the sign of his monoma of higher degree.)
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