Question

What are the global and local extrema of `f(x)=8x^3-4x^2+6` ?

Answer 1

The local extrema are `(0,6)` and `(1/3,158/27)`
and the global extrema are `+-oo`

Explanation:

We use `(x^n)'=nx^(n-1)`
Let us find the first derivative

`f'(x)=24x^2-8x`
For local extrema `f'(x)=0`
So `24x^2-8x=8x(3x-1)=0`

`x=0` and `x=1/3`

So let's do a chart of signs
`x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `0` `color(white)(aaaaa)` `1/3` `color(white)(aaaaa)` `+oo`
`f'(x)` `color(white)(aaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+`
`f(x)` `color(white)(aaaaaa)` `uarr` `color(white)(aaaaa)` `darr` `color(white)(aaaaa)` `uarr`

So at the point `(0,6)` we have a local maximum
and at `(1/3,158/27)`
We have a point a point of inflexion `f''(x)=48x-8`
`48x-8=0` `=>` `x=1/6`
limit`f(x)=-oo`
`xrarr-oo`
limit`f(x)=+oo`
`xrarr+oo`
graph{8x^3-4x^2+6 [-2.804, 3.19, 4.285, 7.28]}