# Question 31098

Nov 7, 2017

At the inflection point $x = 2 - \sqrt{2}$, the slope is:

$\left(2 \sqrt{2} - 2\right) \left({e}^{- 2 + \sqrt{2}}\right)$

At the inflection point $x = 2 + \sqrt{2}$, the slope is:

$\left(- 2 \sqrt{2} - 2\right) \left({e}^{- 2 - \sqrt{2}}\right)$

#### Explanation:

First, let's find $\frac{\mathrm{dy}}{\mathrm{dx}}$ and $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{-} x\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \left(- {e}^{-} x\right) + {e}^{-} x \left(2 x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x - {x}^{2}\right) \left({e}^{-} x\right)$

Now, we derive this again to get the second derivative.

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(2 x - {x}^{2}\right) \left({e}^{-} x\right)$

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

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(2 - 2 x - \left(2 x - {x}^{2}\right)\right) \left({e}^{-} x\right)$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(2 - 4 x + {x}^{2}\right) \left({e}^{-} x\right)$

To find an inflection point, we set the second derivative equal to 0.

$0 = \left({x}^{2} - 4 x + 2\right) \left({e}^{-} x\right)$

We know that ${e}^{-} x$ will never be $0$, so we can divide both sides of the equation by it.

$0 = {x}^{2} - 4 x + 2$

$0 = {x}^{2} - 4 x + 4 - 2$

$0 = {\left(x - 2\right)}^{2} - 2$

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

$\pm \sqrt{2} = x - 2$

$2 \pm \sqrt{2} = x$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we have our two inflection points: $\textcolor{red}{2 - \sqrt{2}}$ and $\textcolor{b l u e}{2 + \sqrt{2}}$

To find the slope of the curve at each of these points, just plug these values into the first derivative formula we calculated at the beginning. Let's start with the first point:

dy/dx]_color(red)(2-sqrt2) = (2(color(red)(2-sqrt2)) - (color(red)(2-sqrt2))^2)(e^-(color(red)(2-sqrt2)))

dy/dx]_color(red)(2-sqrt2) = (4 - 2sqrt2 - (4 - 4sqrt2 + 2))(e^(sqrt2-2))

dy/dx]_color(red)(2-sqrt2) = (2sqrt2 - 2)(e^(-2+sqrt2))

Finally, we plug in the value for the other point:

dy/dx]_color(blue)(2+sqrt2) = (2(color(blue)(2+sqrt2))-(color(blue)(2+sqrt2))^2)(e^-(color(blue)(2+sqrt2)))

dy/dx]_color(blue)(2+sqrt2) = (4+2sqrt2 - (4 + 4sqrt2 + 2))(e^(-2-sqrt2))

dy/dx]_color(blue)(2+sqrt2) = (-2sqrt2 - 2)(e^(-2-sqrt2))#

So these are our two slope values at the two inflection points.