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

Jan 4, 2017

As viewed from O, concave $\left(y ' \uparrow\right)$ for $x < 1$ and concave $\left(y ' \downarrow\right)$ for$x > 1. \left(1 , - 8\right)$ is the point of inflexion (POI), at which the tangent crosses the curve , for reversing rotation.

#### Explanation:

$y = f \left(x\right) = - {x}^{3} + 3 {x}^{2} + 2 x - 12$

$y ' = - 3 {x}^{2} + 6 x + 2 = 0$, at the turning points$x = 1 \pm \sqrt{\frac{5}{3}} = - 0.291 \mathmr{and} 2.291$, nearly

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

y'''=-6 $\ne 0$

So, x =1 gives the point of inflexion (POI) $\left(1. - 8\right)$.

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

anticlockwise to clockwise.

The second graph, the zooming is to see $P O I \left(1. - 8\right)$ 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]} .