Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=2x^3+3x^2-12x`?

Answer 1

The intervals of increasing are `x in ] -oo,-2 [ uu ] -7, oo[`
The interval of decreasing is `[-2,1]`

Explanation:

We need

`(x^n)'=nx^(n-1)`

We calculate the derivative

`y=2x^3+3x^2-12x`

`dy/dx=6x^2+6x-12`

To find the critical points, let `dy/dx=0`

`6x^2+6x-12=6(x^2+x-2)=0`

`6(x-1)(x+2)=0`

Therefore, `x=1` and `x=-2`

We can make a variation table

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaaaaaa)` `1` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `dy/dx` `color(white)(aaaaaaa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaa)` `↗` `color(white)(aa)` `20` `color(white)(aa)` `↘` `color(white)(a)` `-7` `color(white)(aaa)` `↗`

The intervals of increasing are `x in ] -oo,-2 [ uu ] -7, oo[`

The interval of decreasing is `[-2,1]`