# How do you find the critical points for f(x)= 6x^(2/3) + x^(5/3)?

Jan 12, 2016

#### Answer:

$x = - \frac{12}{5} , 0$

#### Explanation:

Use the power rule.

$f ' \left(x\right) = \frac{2}{3} \left(6 {x}^{- \frac{1}{3}}\right) + \frac{5}{3} {x}^{\frac{2}{3}}$

$f ' \left(x\right) = \frac{4}{x} ^ \left(\frac{1}{3}\right) + \frac{5}{3} {x}^{\frac{2}{3}}$

$f ' \left(x\right) = \frac{4}{x} ^ \left(\frac{1}{3}\right) + \frac{5 x}{3 {x}^{\frac{1}{3}}}$

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

Notice that writing the derivative in this form makes finding critical values much simpler.

Critical values occur when the derivative either equals zero (the numerator equals zero) or when the derivative does not exist (the denominator equals zero).

$12 + 5 x = 0$

$x = - \frac{12}{5}$

This is a critical value.

$3 {x}^{\frac{1}{3}} = 0$

$x = 0$

This is the other critical value.

The function graphed:

graph{6x^(2/3) + x^(5/3) [-25.58, 32.16, -10.9, 17.97]}

Notice the maximum at $x = - \frac{12}{5}$ and sharp turn when $x = 0$.