# How do you find the critical points for -2/9x^3 -2/3x^2+16/3x+160/9 and the local max and min?

Sep 28, 2016

See below.

#### Explanation:

$f \left(x\right) = - \frac{2}{9} {x}^{3} - \frac{2}{3} {x}^{2} + \frac{16}{3} x + \frac{160}{9}$

First, let's clean the expression up by factoring out $- \frac{2}{9}$.

$f \left(x\right) = - \frac{2}{9} \left({x}^{3} + 3 {x}^{2} - 24 x - 80\right)$.

Now, we'll find the critical numbers by finding the numbers in the domain at which the derivative is $0$ or fails to exist.

The domain of $f$ is all real numbers.

$f ' \left(x\right) = - \frac{2}{9} \left(3 {x}^{2} + 6 x - 24\right)$.

The derivative exists for all real $x$, so now we need to solve $f ' \left(x\right) = 0$ to find the critical numbers.

$- \frac{2}{9} \left(3 {x}^{2} + 6 x - 24\right) = 0$ is equivalent to

$- \frac{2}{3} \left({x}^{2} + 2 x - 8\right) = 0$

$- \frac{2}{3} \left(x + 4\right) \left(x - 2\right) = 0$

$x = - 4$ $\text{ }$ or $\text{ }$ $x = 2$.

These are the critical numbers for $f$. They determine three intervals on the number line.

We look at the sign of $f '$ on each interval to determine whether $f$ is increasing or decreasing on the interval

{: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"), ((-oo,-4)," " -" ", " "" Decr"), ((-4,2), " " +, " " " Incr"), ((2 ,oo), " " -, " "" Decr") :}

$f$ has a local minimum of $0$ at $- 4$, and

$f$ has a local maximum of 24 at $2$.

Calculations

$f \left(- 4\right) = - \frac{2}{9} \left[{\left(- 4\right)}^{3} + 3 {\left(- 4\right)}^{2} - 24 \left(- 4\right) - 80\right]$

$= - \frac{2}{9} \left[- 4 \left(16\right) + 3 \left(16\right) + 6 \left(16\right) - 5 \left(16\right)\right]$

$= - \frac{2}{9} \left[0 \left(16\right)\right] = 0$

and

$f \left(2\right) = - \frac{2}{9} \left[{\left(2\right)}^{3} + 3 {\left(2\right)}^{2} - 24 \left(2\right) - 80\right]$

$= - \frac{2}{9} \left[2 \left(4\right) + 3 \left(4\right) - 12 \left(4\right) - 20 \left(4\right)\right]$

$= - \frac{2}{9} \left[- 27 \left(4\right)\right] = \frac{2 {\cancel{\left(27\right)}}^{3} \left(4\right)}{\cancel{9}} = 24$