What is the inflection point of ##y=xe^x##?

We need to find where the concavity changes. These are the inflection points; usually it's where the second derivative is zero.

Our function is ##y = f(x) = x e^x##.

Let's see where ##f''(x) = 0##:

##y = f(x) = x*e^x##

So use the product rule:

##f'(x) = x*d/dx(e^x) + e^x*d/dx(x) = x e^x + e^x*1 = e^x(x+1)##

##f''(x) = (x+1)*d/dx(e^x) + e^x*d/dx(x+1)##

## = (x+1) e^x + e^x*1 = e^x(x+2) =0 ##

Set f''(x) = 0 and solve to get x = -2. The second derivative changes sign at -2, and so the concavity changes at x = -2 from concave down to the left of -2 to concave up to the right of -2.

The inflection point is at (x,y) = (-2, f(-2)).

\dansmath leaves it to you to find the y-coordinate!/