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

Feb 15, 2017

Point of inflection is at $x = - 2$ and while in range $\left(- \infty , - 2\right)$ the curve is concave down and in range $\left(- 2 , \infty\right)$ the curve is concave up.

#### Explanation:

At points of inflection, second derivative of the function is equal to zero. Hence le us first wok out second derivative for $y = x {e}^{x}$.

As $y = x {e}^{x}$ and

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \times \frac{d}{\mathrm{dx}} {e}^{x} + 1 \times {e}^{x} = x {e}^{x} + {e}^{x} = {e}^{x} \left(x + 1\right)$

and ${d}^{2} / {\left(\mathrm{dx}\right)}^{2} {e}^{x} \left(x + 1\right)$

= ${e}^{x} \left(x + 1\right) + {e}^{x} = {e}^{x} \left(x + 2\right)$

and this is zero at $x = - 2$

As there is only one point of inflection, we have two ranges $\left(- \infty , - 2\right)$ and (-2,oo)#

In the range $\left(- \infty , - 2\right)$, second derivative is negative and hence in this range the curve is concave down and

in the range $\left(- 2 , \infty\right)$, second derivative is positive and hence in this range the curve is concave up.
graph{xe^x [-7.4, 2.6, -1.56, 3.44]}