Question

How do you find the relative extrema of `f(x)=3x^5-5x^3`?

Answer 1

Please see the explanation below

Explanation:

The function is

`f(x)=3x^5-5x^3`

Calculate the first derivative

`f'(x)=15x^4-15x^2=15x^2(x^2-1)`

The critical points are when `f'(x)=0`

That is,

`15x^2(x^2-1)=0`

The solutions to this equation are

`{(x=0),(x=-1),(x=1):}`

Let's build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaa)` `0` `color(white)(aaaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `↗` `color(white)(aaaa)` `↘` `color(white)(aaaa)` `↘` `color(white)(aaaa)` `↗`

When `x=-1`, there is a relative maximum at `(-1, 2)`

When `x=1`, there is a relative minimum at `(1, -2)`

When `x=0`, there is an inflection point at `(0, 0)`

graph{3x^5-5x^3 [-10, 10, -5, 5]}