# What are the local extrema, if any, of f (x) =(x^3 + 2x^2)/(3 - 5x)?

Nov 22, 2016

Local Extrema:

$x \approx - 1.15$

$x = 0$

$x \approx 1.05$

#### Explanation:

Find the derivative $f ' \left(x\right)$

Set $f ' \left(x\right) = 0$

These are your critical values and potential local extrema.

Draw a number line with these values.

Plug in values within each interval;

if $f ' \left(x\right) > 0$, the function is increasing.

if $f ' \left(x\right) < 0$, the function is decreasing.

When the function changes from negative to positive and is continuous at that point, there is a local minimum; and vice versa.

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

$f ' \left(x\right) = \frac{9 {x}^{2} - 15 {x}^{3} + 12 x - 20 {x}^{2} + 5 {x}^{3} + 10 {x}^{2}}{3 - 5 x} ^ 2$

$f ' \left(x\right) = \frac{- 10 {x}^{3} - {x}^{2} + 12 x}{3 - 5 x} ^ 2$

$f ' \left(x\right) = \frac{- x \left(10 {x}^{2} + x - 12\right)}{3 - 5 x} ^ 2$

Critical values:

$x = 0$

$x = \frac{\sqrt{481} - 1}{20} \approx 1.05$
$x = - \frac{\sqrt{481} + 1}{20} \approx - 1.15$

$x \ne \frac{3}{5}$

<------$\left(- 1.15\right)$------$\left(0\right)$-----$\left(\frac{3}{5}\right)$-----$\left(1.05\right)$------>

Plug in values between these intervals:

You will get a:
Positive value on $\left(- \infty , - 1.15\right)$
Negative on $\left(- 1.15 , 0\right)$
Positive on $\left(0 , \frac{3}{5}\right)$
Positive on $\left(\frac{3}{5} , 1.05\right)$
Negative on $\left(1.05 , \infty\right)$

$\therefore$ Your local maximums will be when:

$x = - 1.15 \mathmr{and} x = 1.05$

Your local minimum will be when:

$x = 0$