Question

How to find the coordinates of the stationary points on the curve `y = x^3 – 6x^2 – 36x + 16`?

Answer 1

We have a local maximum at `(-2,56)` and a local minimum at `(6,-200)` and an inflexion point at `(2,-72)`

Explanation:

We start with the first derivative

`y=x^3-6x^2-36x+16`

`dy/dx=3x^2-12x-36`

The critical points are when `dy/dx=0`

That is,

`3x^2-12x-36=0`

`2(x^2-4x-12)=0`

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

Therefore,

`x=-2` and `x=6`

We build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `6` `color(white)(aaaa)` `+oo`

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

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

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

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

Now, we calculate the second derivative

`(d^2y)/dx^2=6x-12`

We have an inflexion point when, `(d^2y)/dx^2=0`

That is, `x=2`

We make a second chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `]-oo,2[` `color(white)(aaaa)` `]2,+oo[`

`color(white)(aaaa)` `(d^2y)/dx^2` `color(white)(aaaaaaaaaa)` `-` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaaaaaaa)` `nn` `color(white)(aaaaaaaa)` `uu`

We have a local maximum at `(-2,56)` and a local minimum at `(6,-200)` and an inflexion point at `(2,-72)`