Question

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`?

Answer 1

The point of inflexion is at `(-3,4)`
Maximum is at `(-4,6)`
Minimum is at `(-2,2)`

Explanation:

We start by calculating the derivatives.
`y=x^3+9x^2+24x+22`
Since this is a polynomial function with no denominators, there are no discontinuities. The curve is continuous.
`dy/dx=3x^2+18x+24`
For maximum and minimum `dy/dx=0`
So `3x^2+18x+24=0` `=>` `x^2+6x+8=0`
`=>` `(x+2)(x+4)=0` so `x=-2` and `x=-4`
Between `-oo and -4` `dy/dx` is positive and the curve is increasing
Between `-4 and -2` `dy/dx` is negative and the curve is decreasing
Between `-2and +oo` `dy/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 `(d^2y)/dx^2`
So `(d^2y)/dx^2=6x+18` `=>` `x=-3`

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