Question

Is `f(x) =-x^3-(x-2)(x-1)` concave or convex at `x=-1`?

Answer 1

The function is convex.

Explanation:

We calculate the first and second derivatives.

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

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

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

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

`f''(x)=-6x-2`

There is a point of inflexion when `f''(x)=0`

`-6x-2=0`

`6x=-2`

`x=-2/6=-1/3`

We build a chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,-1/3)` `color(white)(aaaa)` `(-1/3,+oo)`

`color(white)(aaaa)` `f''(x)` `color(white)(aaaaaaaaaaaa)` `+` `color(white)(aaaaaaaaaaaaa)` `-`

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

At `x=-1`

`f''(-1)=(-6*-1)-2=4`

As `f''(x)>0`, the function `f(x)` is convex.

Also,

`(-1) in (-oo,-1/3)` and the curve is `uu` (convex)

graph{-x^3-(x-2)(x-1) [-12.66, 12.65, -6.33, 6.33]}