Question

What are the values and types of the critical points, if any, of `f(x)=x^3 + 3x^2 + 1`?

Answer 1

We have a local maximum at `(-2,5)`
We have a local minimum at `(0,1)`
The inflexion point is `(-1,3)`

Explanation:

We calculate the first derivative

`f(x)=x^3+3x^2+1`

`f'(x)=3x^2+6x`

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

`3x^2+6x=0`

Factorising yields

`3x(x+2)=0`

Therefore,

`x=0 or x=-2`

We build a chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+oo`

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

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

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

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

Now, we calculate the second derivative

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

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

`6x+6=0`

`x=-1`

The inflexion point is `(-1,3)`

We calculate

`f''(-2)=6*-2+6=-6`

As `f''(-2)<0`, we have a local maximum at `(-2,5)`

`f''(0)=6`

As `f''(0)>0`, we have a local minimum at `(0,1)`

We build another chart to determine the convexity and concavity

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,-1)` `color(white)(aaaa)` `(-1,+oo)`

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

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaaaa)` `nn` `color(white)(aaaaaaaaaaa)` `uu`

graph{x^3+3x^2+1 [-14.66, 13.82, -5.92, 8.32]}