# What are the points of inflection, if any, of f(x)=xe^(-x^2) ?

Jun 1, 2017

The points of inflection are at $- 1.2247 , 0 , \mathmr{and} 1.2247$.

#### Explanation:

To find points of inflection for a function, you take the second derivative of the function and set the second derivative to equal 0. The product rule is used below:

$f \left(x\right) = x {e}^{- {x}^{2}}$

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

${f}^{\left(2\right)} \left(x\right) = \left(- 4 x \cdot {e}^{- {x}^{2}}\right) + \left(- 2 {x}^{2} \cdot - 2 x {e}^{- {x}^{2}}\right) - 2 x {e}^{- {x}^{2}}$

${f}^{\left(2\right)} \left(x\right) = - 4 x {e}^{- {x}^{2}} + 4 {x}^{3} {e}^{- {x}^{2}} - 2 x {e}^{- {x}^{2}}$

${f}^{\left(2\right)} \left(x\right) = - 6 x {e}^{- {x}^{2}} + 4 {x}^{3} {e}^{- {x}^{2}}$

Now we set it equal to 0:

$0 = - 6 x {e}^{- {x}^{2}} + 4 {x}^{3} {e}^{- {x}^{2}}$

If we graph this function, the roots are approximately at $- 1.2247 , 0 , \mathmr{and} 1.2247$:
graph{-6xe^(-x^2)+4x^3e^(-x^2) [-5, 5, -2.5, 2.5]}

All of these values go from positive to negative or vice versa in an interval test, so they are all points of inflection.