Question

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

Answer 1

Local Extrema:

`x~~-1.15`

`x=0`

`x~~1.05`

Explanation:

Find the derivative `f'(x)`

Set `f'(x)=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'(x) >0`, the function is increasing.

if `f'(x) <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'(x)=[(3x^2+4x)(3-5x)-(-5)(x^3+2x^2)]/(3-5x)^2`

`f'(x)=[9x^2-15x^3+12x-20x^2+5x^3+10x^2]/(3-5x)^2`

`f'(x)=(-10x^3-x^2+12x)/(3-5x)^2`

`f'(x)=[-x(10x^2+x-12)]/(3-5x)^2`

Critical values:

`x=0`

`x=(sqrt(481)-1)/20~~1.05`
`x=-(sqrt(481)+1)/20~~-1.15`

`x!=3/5`

<------`(-1.15)`------`(0)`-----`(3/5)`-----`(1.05)`------>

Plug in values between these intervals:

You will get a:
Positive value on `(-oo, -1.15)`
Negative on `(-1.15, 0)`
Positive on `(0, 3/5)`
Positive on `(3/5, 1.05)`
Negative on `(1.05, oo)`

`:.` Your local maximums will be when:

`x=-1.15 and x=1.05`

Your local minimum will be when:

`x=0`