How do you determine the interval(s) where the function $f (x) = x e^{- x}$ $f(x) = xe^-x$ is concave down?

Question

5540 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-30T07:55:57

A function is concave down in the intervals in which its second derivative is negative.

So:

$y'=1*e^-x+xe^-x*(-1)=e^-x-xe^-x$

$y''=e^-x*(-1)-[1*e^-x+xe^-x*(-1)]=$

$=-e^-x-e^-x+xe^-x=xe^-x-2e^-x=$

$=e^-x(x-2)$

$y''<0rArrx<2$ .

(this is because the exponential function is always positive)

It's clear also seeing the graph:

graph{xe^-x [-10, 10, -5, 5]}

The point $A(2,2e^-2)$ is an inflection point.

How do you determine the interval(s) where the function f(x) = xe^-xf(x)=xe−xf(x) = xe^-x is concave down?