# What are the critical values, if any, of f(x)= x^(3/4) - 2x^(1/4)?

Feb 14, 2016

By the definition I am accustomed to, they are $0$ and $\frac{4}{9}$.

#### Explanation:

A critical value of $f$ is a value in the domain of $f$ at which $f '$ does not exists or $f ' \left(x\right)$ is $0$.

The domain of the function $f \left(x\right) = {x}^{\frac{3}{4}} - 2 {x}^{\frac{1}{4}}$ is $\left[0 , \infty\right)$.

The derivative is $f ' \left(x\right) = \frac{3}{4} {x}^{- \frac{1}{4}} - \frac{1}{2} {x}^{- \frac{3}{4}} = \frac{3 {x}^{\frac{1}{2}} - 2}{4 {x}^{\frac{3}{2}}}$

$f '$ fails to exist at $x = 0$ and $f ' \left(x\right) = 0$ at $x = \frac{4}{9}$.

Both $0$ and $\frac{4}{9}$ are in the domain, so both are critical values.