Question

How do you find the intervals of increasing and decreasing given `y=x^3-11x^2+39x-47`?

Answer 1

The intervals of increasing are `]-oo,3[` and `]13/3,+oo[`
The interval of decreasing is `]3,13/3[`

Explanation:

We calculate the first derivative

`y=x^3-11x^2+39x-47`

`dy/dx=3x^2-22x+39`

We need `dy/dx=0`

`3x^2-22x+39=(3x-13)(x-3)=0`

The critical points are

`x=13/3` and `x=3`

Now, we construct the chart

`color(white)(aa)` `Interval` `color(white)(aa)` `|` `color(white)(aaa)` `]-oo,3[` `color(white)(aaa)` `|` `color(white)(aaa)` `]3,13/3[` `color(white)(aaa)` `|` `]13/3,+oo[`

`color(white)(aa)` `x-3` `color(white)(aaaaa)` `|` `color(white)(aaa)` `-` `color(white)(aaaaaaa)` `|` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `|` `color(white)(aaaa)` `+`

`color(white)(aa)` `3x-13` `color(white)(aaa)` `|` `color(white)(aaa)` `-` `color(white)(aaaaaaa)` `|` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `|` `color(white)(aaaa)` `+`

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

`color(white)(aa)` `y` `color(white)(aaaaaaaaa)` `|` `color(white)(aaa)` `↗` `color(white)(aaaaaa)` `|` `color(white)(aaaaaa)` `↘` `color(white)(aaaa)` `|` `color(white)(aaaa)` `↗`

The intervals of increasing are `]-oo,3[` and `]13/3,+oo[`

The interval of decreasing is `]3,13/3[`