Question

How do you find the local maximum and minimum values of `f(x)=x^3 + 6x^2 + 12x -1` using both the First and Second Derivative Tests?

Answer 1

There are no minimum or maximum, only a point of inflection at `(-2,-9)`

Explanation:

Let's calculate the first and second derivatives

`f(x)=x^3+6x^2+12x-1`

`f'(x)=3x^2+12x+12`

`=3(x^2+4x+4)`

`=3(x+2)(x+2)`

`=3(x+2)^2`

and

`f''(x)=6x+12`

The critical point is when `f'(x)=0`

That is,

`x+2=0`

`x=-2`

Let's build a chart

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

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `↗` `color(white)(aaaaaa)` `↗`

Let's look at `f''(x)`

`f''(x)=0` when `x+2=0`

That is when `x=-2`

There is a point of inflection at `(-2,-9)`

Let's do a chart

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

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

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `nn` `color(white)(aaaaaa)` `uu` graph{x^3+6x^2+12x-1 [-38.27, 26.68, -22.07, 10.37]}