Question

For what values of x is `f(x)= -x^4-4x^3+8x^2+6x+2` concave or convex?

Answer 1

Convex ( f'`uarr`) towards x-axis for `x in (-1-1/sqrt3, -1+1/sqrt3)`, These are points of inflexion (`f'' = 0 and f''' ne 0`).
Elsewhere, the graph is concave ( f' `darr` ).

Explanation:

The second derivative

`f''=-4(3x^2+6x-4)=-0`, `at x =-1+-1/sqrt3= -1.577 and -.4226`,

nearly. f'''=-24x-24 < 0, at these points.

So, they are points of inflexion.

In this interval, f > 0. See the second graph for clarity.

The first reveals the turning points. The middle one is well above x-

axis.

The points of inflexion are about this point.

The third if f'- graph

The ad hoc x and y scales befit clarification.

ph{-x^4-4x^3+8x^2+6x+2 [-16, 16, -128, 128]}

graph{-x^4-4x^3+8x^2+6x+2 [,-1 1, -.5, .5]}

graph{-4x^3-12x^2+16x+6 [-5, 5 -2.5, 2.5]}