# How do you find the inflection points of the graph of the function: f(x) = xe^(-2x)?

Jun 6, 2015

The inflection points are given when the second derivative equals zero, that is, the function is changing its concavity (from concave to convex or vice-versa).

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- 2 x} - 2 x {e}^{- 2 x} = \textcolor{g r e e n}{{e}^{- 2 x} \left(1 - 2 x\right)}$

(d^2y)/(dx^2)=-2e^(-2x)(1-2x)+e^(-2x)(-2)=-2e^(-2x)((color(blue)(1)-2x+color(blue)(1))

$\frac{{d}^{2} y}{\mathrm{dx}} = - 2 {e}^{- 2 x} \left(\textcolor{b l u e}{2} - 2 x\right) = - 2 {e}^{- 2 x} \left[\textcolor{b l u e}{2} \left(1 - x\right)\right] = \textcolor{red}{- 4 {e}^{- 2 x} \left(1 - x\right)}$

Now, let's just rewrite the function as a quotient: $- \frac{4 \left(1 - x\right)}{e} ^ \left(2 x\right)$

The inflection point is given by the point where the second derivative equals zero, so, when will it be zero, as the denominator will never be exactly zero? The answer is: when the numerator becomes zero, which is clear to see: when the factor $\left(1 - x\right)$ equals zero, the whole function is zero; thus, when $x = 1$, it all becomes zero.

Substituting the value of $x = 1$ in the original function:

$f \left(1\right) = \left(1\right) {e}^{- 2 \left(1\right)} = \frac{1}{e} ^ 2 \cong 0.1353$

The inflection point, thus, is $\left(1 , 0.1353\right)$