# How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for y=x^3+9x^2+24x+22?

Oct 19, 2016

The point of inflexion is at $\left(- 3 , 4\right)$
Maximum is at $\left(- 4 , 6\right)$
Minimum is at $\left(- 2 , 2\right)$

#### Explanation:

We start by calculating the derivatives.
$y = {x}^{3} + 9 {x}^{2} + 24 x + 22$
Since this is a polynomial function with no denominators, there are no discontinuities. The curve is continuous.
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + 18 x + 24$
For maximum and minimum $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
So $3 {x}^{2} + 18 x + 24 = 0$ $\implies$ ${x}^{2} + 6 x + 8 = 0$
$\implies$ $\left(x + 2\right) \left(x + 4\right) = 0$ so $x = - 2$ and $x = - 4$
Between $- \infty \mathmr{and} - 4$ $\frac{\mathrm{dy}}{\mathrm{dx}}$ is positive and the curve is increasing
Between $- 4 \mathmr{and} - 2$ $\frac{\mathrm{dy}}{\mathrm{dx}}$ is negative and the curve is decreasing
Between $- 2 \mathmr{and} + \infty$ $\frac{\mathrm{dy}}{\mathrm{dx}}$ is positive and the curve is increasing
So we have a maximum when $x = - 4$
and a minimum when $x = - 2$
to find points of inflexions, we calculate $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$
So $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 x + 18$ $\implies$ $x = - 3$

graph{x^3+9x^2+24x+22 [-20, 20, -10, 10]}