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

##### 1 Answer
Dec 1, 2015

Critical value is $x = \frac{1}{2}$

#### Explanation:

$f \left(x\right) = {x}^{\frac{2}{3}} + {x}^{- \frac{1}{3}}$

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

To find the critical points set $f ' \left(x\right) = 0$, and solve for $x$

$\frac{2 {x}^{- \frac{1}{3}} - {x}^{- \frac{4}{3}}}{3} = 0$

$2 {x}^{- \frac{1}{3}} - {x}^{- \frac{4}{3}} = 0$

$2 {x}^{- \frac{1}{3}} = {x}^{- \frac{4}{3}} \to$ use law of exponential

$\frac{2}{{x}^{\frac{1}{3}}} = \frac{1}{{x}^{\frac{4}{3}}} \to$ cross multiplication

${x}^{\frac{4}{3}} / {x}^{\frac{1}{3}} = \frac{1}{2}$

$x = \frac{1}{2}$