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

Question

Scarlett Allison · Accepted 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 #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!/

Evelyn Agard · Answer

undefined To find the inflection point of $ y = xe^x $, we need to find the second derivative and then set it equal to zero.

First, the first derivative of $ y = xe^x $ is $ y' = e^x + xe^x $.

Then, the second derivative is $ y'' = e^x + xe^x + e^x = 2e^x + xe^x $.

Setting $ y'' $ equal to zero and solving for $ x $, we get:

$ 2e^x + xe^x = 0 $

$ e^x(2 + x) = 0 $

This equation is equal to zero when $ e^x = 0 $ (which is not possible) or when $ 2 + x = 0 $.

Solving $ 2 + x = 0 $, we get $ x = -2 $.

Therefore, the inflection point of $ y = xe^x $ is at $ (x, y) = (-2, -2e^{-2}) $.