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

##### 1 Answer
Jun 8, 2017

At $x \cong 0.8$

#### Explanation:

Point of inflection occur where $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 0$

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

$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{{x}^{2}} + x {e}^{{x}^{2}} \times 2 x - 2 x {e}^{x} - {x}^{2} {e}^{x}$

= ${e}^{{x}^{2}} + 2 {x}^{2} {e}^{{x}^{2}} - 2 x {e}^{x} - {x}^{2} {e}^{x}$

and $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 2 x {e}^{{x}^{2}} + 2 \left(2 x {e}^{{x}^{2}} + {x}^{2} \times {e}^{{x}^{2}} \times 2 x\right) - 2 \left({e}^{x} + x {e}^{x}\right) - \left(2 x {e}^{x} + {x}^{2} {e}^{x}\right)$

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

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

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

at $x \cong 0.8$ (from the graph given below) and this is the point of inflection, as may be seen from graph of $f \left(x\right) = x {e}^{{x}^{2}} - {x}^{2} {e}^{x}$

graph{6xe^(x^2)+4x^3e^(x^2)-2e^x-4xe^x-x^2e^x [-0.553, 1.947, -0.545, 0.705]}

graph{xe^(x^2)-x^2e^x [-0.553, 1.947, -0.545, 0.705]}