Question

How do you find the intervals of increasing and decreasing given `y=(3x^2-3)/x^3`?

Answer 1

The intervals of decreasing are `x in ] -oo,-sqrt3 ] uu [sqrt3, oo[`

The intervals of increasing are `x in [ -sqrt3,0 [ uu ] 0, sqrt3 ]`

Explanation:

The domain of `y` is `D_y=RR-{0}`

We calculate the derivative to find the intervals of increasing and decreasing.

Here, we have a fraction of 2 functions

`(u/v)'=(u'v-uv')/(v^2)` and `(x^n)'=nx^(n-1)`

`u=3x^2-3`, `=>` `u'=6x`

`v=x^3`, `=>`, `v'=3x^2`

So,

`dy/dx=(6x^4-9x^4+9x^2)/(x^6)`

`=(9x^2-3x^4)/(x^6)=(3x^2(3-x^2))/(x^6)`

`=(3(3-x^2))/x^4`

So,

`dy/dx=0`, when `3-x^2=0`

`x=-sqrt3` and `x=sqrt3`

Now, we can make our sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-sqrt3` `color(white)(aaaa)` `0` `color(white)(aaaa)` `sqrt3` `color(white)(aaaa)` `+oo`

`color(white)(aa)` `dy/dx` `color(white)(aaaaaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aa)` `∣∣` `color(white)(aa)` `+` `color(white)(aaa)` `-`

`color(white)(aa)` `y` `color(white)(aaaaaaaaaa)` `darr` `color(white)(aaaaa)` `uarr` `color(white)(aaa)` `∣∣` `color(white)(aa)` `uarr` `color(white)(aaa)` `darr`

So,
The intervals of decreasing are `x in ] -oo,-sqrt3 ] uu [sqrt3, oo[`

The intervals of increasing are `x in [ -sqrt3,0 [ uu ] 0, sqrt3 ]`

graph{(3x^2-3)/(x^3) [-10, 10, -5, 5]}