Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=(x^5-4x^3+3x)/(x^2-1)`?

Answer 1

The intervals of increasing are `x in (-oo,-1) uu (1,+oo)`
The interval of decreasing is `x in (-1,1)`

Explanation:

We need

`(u/v)'=(u'v-uv')/(v^2)`

Our function is

`y=(x^5-4x^3+3x)/(x^2-1)`

`u(x)=x^5-4x^3+3x`, `=>`, `u'(x)=5x^4-12x^2+3`

`v(x)=x^2-1`, `=>`, `v'(x)=2x`

Therefore,

`dy/dx=((5x^4-12x^2+3)(x^2-1)-(x^5-4x^3+3x)(2x))/(x^2-1)^2`

`=((5x^4-12x^2+3)(x^2-1)-x(x^2-1)(x^2-3)(2x))/(x^2-1)^2`

`=((x^2-1)(5x^4-12x^2+3-2x^4+6x^2))/(x^2-1)^2`

`=(3x^4-6x^2+3)/(x^2-1)`

`=(3(x^4-2x^2+1))/(x^2-1)`

`=(3(x^2-1)^2)/(x^2-1)`

`=3(x^2-1)`

The critical points are when `dy/dx=0`

`3(x^2-1)=3(x+1)(x-1)=0`, `=>`, `x=1` and `x=-1`

We can build the variation chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,-1)` `color(white)(aaaa)` `(-1,1)` `color(white)(aaaa)` `(1,+oo)`

`color(white)(aaaa)` `Sign dy/dx` `color(white)(aaaaaaaaa)` `+` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaaaaaaaa)` `↗` `color(white)(aaaaaaaaaa)` `↘` `color(white)(aaaaaaaa)` `↗`

The intervals of increasing are `x in (-oo,-1) uu (1,+oo)`

The interval of decreasing is `x in (-1,1)`

We could have obtained the same result by simplifying `y`

`y=(x^5--4x^3+3x)/(x^2-1)=(x(x^2-3)(x^2-1))/(x^2-1)=x^3-3x`

`dy/dx=3x^2-3=3(x^2-1)=3(x+1)(x-1)`

graph{(x^5-4x^3+3x)/(x^2-1) [-10, 10, -5, 5]}