# What are the critical points of f(x) = 6x^(2/3) + x^(5/3)?

Oct 27, 2015

Thee critical numbers for $f$ are $- \frac{12}{5}$ and $0$..

#### Explanation:

For $f \left(x\right) = 6 {x}^{\frac{2}{3}} + {x}^{\frac{5}{3}}$, we have

$\text{Dom} \left(f\right) = \left(- \infty , \infty\right)$ and

$f ' \left(x\right) = \frac{12}{3} {x}^{- \frac{1}{3}} + \frac{5}{3} {x}^{\frac{2}{3}}$

$= \frac{1}{3} {x}^{- \frac{1}{3}} \left(12 + 5 x\right)$

$= \frac{12 + 5 x}{3 {x}^{\frac{1}{3}}}$.

$f ' \left(x\right)$ is undefined at $x = 0$ and
$f ' \left(x\right) = 0$ at $x = - \frac{5}{12}$.

Both of these numbers are in $\text{Dom} \left(f\right)$, so both are critical numbers for $f$.

Alternative Method

Many students find it more clear to do the algebra differently.

$f ' \left(x\right) = \frac{12}{3} {x}^{- \frac{1}{3}} + \frac{5}{3} {x}^{\frac{2}{3}}$

$= \frac{12}{3 \sqrt[3]{x}} + \frac{5 {\sqrt[3]{x}}^{2}}{3}$

$= \frac{12}{3 \sqrt[3]{x}} + \frac{5 {\sqrt[3]{x}}^{2}}{3} \cdot \frac{\sqrt[3]{x}}{\sqrt[3]{x}}$.

$= \frac{12}{3 \sqrt[3]{x}} + \frac{5 x}{3 \sqrt[3]{x}}$

$= \frac{12 + 5 x}{3 \sqrt[3]{x}}$

Now find $0$ and $- \frac{12}{5}$ as above.