Question

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

Answer 1

Point of inflection is at `x=-2` and while in range `(-oo,-2)` the curve is concave down and in range `(-2,oo)` 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=xe^x`.

As `y=xe^x` and

`(dy)/(dx)=x xx d/(dx)e^x+1xxe^x=xe^x+e^x=e^x(x+1)`

and `d^2/(dx)^2e^x(x+1)`

= `e^x(x+1)+e^x=e^x(x+2)`

and this is zero at `x=-2`

As there is only one point of inflection, we have two ranges `(-oo,-2)` and (-2,oo)#

In the range `(-oo,-2)`, second derivative is negative and hence in this range the curve is concave down and

in the range `(-2,oo)`, 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]}