# How do you find points of inflection and determine the intervals of concavity given y=x^3-2x^2-2?

Jul 23, 2018

The function is concave when $x \in \left(- \infty , \frac{2}{3}\right)$ and convex when $\left(\frac{2}{3} , + \infty\right)$. The point of inflection is at $\left(0.667 , - 2.593\right)$

#### Explanation:

The function is

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

Calculate the first and second derivatives

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 4 x$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 x - 4$

The points of inflection are when

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0$

That is

$6 x - 4 = 0$

$\implies$, $x = \frac{4}{6} = \frac{2}{3}$

The point of inflection is at $\left(0.667 , - 2.593\right)$

The intervals to consider are

${I}_{1} = \left(- \infty , \frac{2}{3}\right)$ and ${I}_{2} = \left(\frac{2}{3} , + \infty\right)$

Let's build a variation chart to determine the concavity

$\textcolor{w h i t e}{a a a a}$$\text{ Interval }$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , \frac{2}{3}\right)$$\textcolor{w h i t e}{a a a a}$$\left(\frac{2}{3} , \infty\right)$

$\textcolor{w h i t e}{a a a a}$$\text{ Sign } \frac{{d}^{2} y}{\mathrm{dx}} ^ 2$$\textcolor{w h i t e}{a a a a a a}$$\left(-\right)$$\textcolor{w h i t e}{a a a a a a a a a}$$\left(+\right)$

$\textcolor{w h i t e}{a a a a}$$\text{ Concavity }$$\textcolor{w h i t e}{a a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a a a a}$$\cup$

graph{x^3-2x^2-2 [-16.02, 16.01, -8.01, 8.01]}