What is the inflection point of #y=xe^x#?
1 Answer
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
Let's see where
#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)#
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!/
