Question

For what values of x is `f(x)= -x^3+3x^2+2x-12` concave or convex?

Answer 1

As viewed from O, concave `( y' uarr)` for `x < 1` and concave `(y' darr)` for`x >1. (1, -8)` is the point of inflexion (POI), at which the tangent crosses the curve , for reversing rotation.

Explanation:

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

`y'=-3x^2+6x+2 =0`, at the turning points`x =1+-sqrt(5/3)=-0.291 and 2.291`, nearly

y''--6(x-1)=0#, at x =1.

y'''=-6 `ne 0`

So, x =1 gives the point of inflexion (POI) `(1. -8)`.

Here, the tangent crosses the curve, reversing rotation, from

anticlockwise to clockwise.

The second graph, the zooming is to see `POI (1. -8)` in `Q_4`,

at the level `y = - 8`.

graph{-x^3+3x^2+2x-12 [-29.95, 29.95, -14.97, 14.98]}

graph{(-x^3+3x^2+2x-12-y)(y+8)=0 [-2, 2, -20, 20]} .