Question

How do you find the maximum, minimum and inflection points and concavity for the function `y=x^4-8x^3`?

Answer 1

Please see the explanation below

Explanation:

First calculate, the first derivative

`y=x^4-8x^3`

`y'=4x^3-24x^2=4x^2(x-6)`

The critical points are when `y'=0`

`4x^2(x-6)=0`

`=>`, `{(x=0),(x=6):}`

Build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `0` `color(white)(aaaaaa)` `6` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x^2` `color(white)(aaaaaaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `x-6` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `y'` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaa)` `↘` `color(white)(aaaa)` `↘` `color(white)(aaaaa)` `↗`

There is a local minimum at `(6, -432)`

Calculate the second derivative

`y''=12x^2-48x=12x(x-4)`

The inflection points are when `y''=0`

`12x(x-4)=0`

`=>`, `{(x=0),(x=4):}`

Build a variation chart

`color(white)(aaaa)` `" Interval "` `color(white)(aaaa)` `(-oo,0)` `color(white)(aaaa)` `(0,4)` `color(white)(aaaaaa)` `(4,+oo)`

`color(white)(aaaaaa)` `"Sign y'' "` `color(white)(aaaaaaa)` `+` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaaaa)` `+`

`color(white)(aaaaaa)` `" y "` `color(white)(aaaaaaaaaaa)` `uu` `color(white)(aaaaaaa)` `nn` `color(white)(aaaaaaaaa)` `uu`

The inflection points are `(0,0)` and `(4,-256)`

The curve is convex when `x in (-oo,0) uu(4, +oo)`

The curve is concave when `x in (0,4)`

graph{x^4-8x^3 [-25.66, 25.67, -12.83, 12.83]}