Question

For what values of x is `f(x)=(x+2)(x-3)(x+1)` concave or convex?

Answer 1

The function is concave for `x in (-oo,0)`
The function is convex for `x in (0,+oo)`

Explanation:

We need

`(uvw)'=u'vw+uv'w+uvw'`

We must calculate the second derivative and determine the sign.

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

`f'(x)=(x-3)(x+1)+(x+2)(x+1)+(x+2)(x-3)`

`=x^2-2x-3+x^2+3x+2+x^2-x-6`

`=3x^2-7`

`f''(x)=6x`

Therefore,

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

This is the point of inflexion.

We can build a chart

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

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

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