# How do you find increasing, decreasing, inflection points, minimum and maximum for the graph f(x) = ln(x)/(8sqrtx)?

Sep 16, 2015

See the explanation.

#### Explanation:

We have to find 1st and 2nd derivative:

$f ' \left(x\right) = \frac{\frac{1}{x} \cdot 8 \sqrt{x} - \frac{8}{2 \sqrt{x}} \ln x}{64 x} = \frac{\frac{1}{\sqrt{x}} - \ln \frac{x}{2 \sqrt{x}}}{8 x} =$

$= \frac{2 - \ln x}{16 x \sqrt{x}}$

${f}^{\left(2\right)} \left(x\right) = \frac{- \frac{1}{x} \cdot 16 x \sqrt{x} - 16 \cdot \frac{3}{2} \cdot \sqrt{x} \cdot \left(2 - \ln x\right)}{{16}^{2} {x}^{3}} =$

$= \frac{- 16 \sqrt{x} - 24 \sqrt{x} \left(2 - \ln x\right)}{{16}^{2} {x}^{3}} =$

$= \frac{- 8 \sqrt{x} \left(2 + 6 - 3 \ln x\right)}{{16}^{2} {x}^{3}} = \frac{- \sqrt{x} \left(8 - 3 \ln x\right)}{32 {x}^{3}}$

Stationary points are zeros of 1st derivative:

$f ' \left({x}_{s}\right) = 0 \iff \frac{2 - \ln {x}_{s}}{16 {x}_{s} {\sqrt{x}}_{s}} = 0 \iff$

$\left(2 - \ln {x}_{s} = 0 \wedge 16 {x}_{s} {\sqrt{x}}_{s} \ne 0\right) \iff$

$\left(\ln {x}_{s} = 2 \wedge {x}_{s} \ne 0\right) \iff \left({x}_{s} = {e}^{2} \wedge {x}_{s} \ne 0\right) \iff {x}_{s} = {e}^{2}$

$\forall x > {x}_{s} : \ln x > 2 \wedge 2 - \ln x < 0 \wedge 16 x \sqrt{x} > 0$

$\forall x > {x}_{s} : f ' \left(x\right) < 0$, $f$ is decreasing

$\forall x < {x}_{s} \wedge x > 0 : \ln x < 2 \wedge 2 - \ln x > 0 \wedge 16 x \sqrt{x} > 0$

$\forall x < {x}_{s} \wedge x > 0 : f ' \left(x\right) > 0$, $f$ is increasing

For $x = {x}_{s}$ function $f$ has maximum value:

${f}_{\max} = f \left({e}^{2}\right) = \ln {e}^{2} / \left(8 \sqrt{{e}^{2}}\right) = \frac{2}{8 e} = \frac{1}{4 e}$

Inflection points are zeros of 2nd derivative if 2nd derivative changes sign in those points and iff function is continuous :

${f}^{\left(2\right)} = 0 \iff \frac{- {\sqrt{x}}_{i} \left(8 - 3 \ln {x}_{i}\right)}{32 {x}_{i}^{3}} = 0 \iff$

$\left(- {\sqrt{x}}_{i} \left(8 - 3 \ln {x}_{i}\right) = 0 \wedge 32 {x}_{i}^{3} \ne 0\right) \iff$

$\left(8 - 3 \ln {x}_{i} = 0 \wedge {x}_{i} \ne 0\right) \iff \left(\ln {x}_{i} = \frac{8}{3} \wedge {x}_{i} \ne 0\right) \iff$

$\left({x}_{i} = {e}^{\frac{8}{3}} \wedge {x}_{i} \ne 0\right) \iff {x}_{i} = {e}^{\frac{8}{3}} = {e}^{2} \sqrt[3]{{e}^{2}}$

So, ${x}_{i} = {e}^{2} \sqrt[3]{{e}^{2}}$ is potential inflection point.

$\forall x > {x}_{i} : 8 - 3 \ln x < 0 , {f}^{\left(2\right)} > 0$
$\forall x < {x}_{i} \wedge x > 0 : 8 - 3 \ln x > 0 , {f}^{\left(2\right)} < 0$

Indeed, ${x}_{i}$ is inflection point:

$f \left({x}_{i}\right) = f \left({e}^{\frac{8}{3}}\right) = \frac{\ln {e}^{\frac{8}{3}}}{8 \sqrt{{e}^{\frac{8}{3}}}} = \frac{\frac{8}{3}}{8 {e}^{\frac{4}{3}}} = \frac{1}{3 e \sqrt[3]{e}}$

Note: $f$ is defined for $\forall x > 0$ because of $\ln x$, so I used that fact.