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

Aug 4, 2014

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 $y = f \left(x\right) = x {e}^{x}$.

Let's see where $f ' ' \left(x\right) = 0$:

$y = f \left(x\right) = x \cdot {e}^{x}$

So use the product rule:

$f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right) + {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right) = x {e}^{x} + {e}^{x} \cdot 1 = {e}^{x} \left(x + 1\right)$

$f ' ' \left(x\right) = \left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right) + {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right)$

$= \left(x + 1\right) {e}^{x} + {e}^{x} \cdot 1 = {e}^{x} \left(x + 2\right) = 0$

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!/