# (a) Find the interval on which f is increasing and decreasing? (b) Find the local maximum value of f? (c) Find the inflection point? (d) Find the interval on which f is concave up and concave down?

Feb 6, 2018

See the explanation below

#### Explanation:

Start by calculating the first derivative, the function $f \left(x\right)$ is the multiplication of $2$ functions.

The domain of $f \left(x\right)$ is $x \in \left[0 , + \infty\right)$

Apply the product rule,

$\left(u v\right) ' = u ' v + u v '$

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

Setting, $f ' \left(x\right) = 0$, gives the critical points

${e}^{-} x \frac{\left(1 - 2 x\right)}{2 \sqrt{x}} = 0$, $\implies$, $x = \frac{1}{2}$

Build a variation chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$\frac{1}{2}$$\textcolor{w h i t e}{a a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$color(white)(a)$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$color(white)(a)$\textcolor{w h i t e}{a a}$↗$\textcolor{w h i t e}{a a}$$0.429$$\textcolor{w h i t e}{a a}$↘

The interval of increasing is $\left(0 , \frac{1}{2}\right)$ and the interval of decreasing is $\left(\frac{1}{2} , + \infty\right)$

The maximum is at the point $\left(0.5 , 0.429\right)$

Calculate the second derivative

$f ' ' \left(x\right) = - {e}^{-} x \frac{1 - 2 x}{2 \sqrt{x}} - {e}^{-} \frac{x}{\sqrt{x}} - {e}^{-} x \frac{1 - 2 x}{4 {x}^{\frac{3}{2}}}$

$= {e}^{-} x \frac{4 {x}^{2} - 4 x - 1}{4 {e}^{\frac{3}{2}}}$

Setting, $f ' ' \left(x\right) = 0$ gives the inflection points.

$4 {x}^{2} - 4 x - 1 = 0$

$x = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot \left(4\right) \cdot \left(- 1\right)}}{2 \cdot 4} = \frac{4 \pm \sqrt{32}}{8}$

${x}_{1} = \frac{1 + \sqrt{2}}{2} = 1.21$

${x}_{2} = \frac{1 - \sqrt{2}}{2} = - 0.21$

The inflection point is $\left(1.21 , 0.328\right)$

Make a variation chart to find the concavities

$\textcolor{w h i t e}{a a a a}$$\text{Interval}$$\textcolor{w h i t e}{a a a a}$$\left(0 , 1.21\right)$$\textcolor{w h i t e}{a a a a}$$\left(1.21 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$\text{Sign f''(x)}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\text{Sign f''(x)}$$\textcolor{w h i t e}{a a a a a}$$\cap$$\textcolor{w h i t e}{a a a a a a a a a}$$\cup$

The interval of concave down is $x \in \left(0 , 1.21\right)$ and the interval of concave up is $x \in \left(1.21 , + \infty\right)$

graph{sqrtx e^-x [-0.821, 3.024, -0.854, 1.068]}