# How do you find the monotonicity, extrema, concavity, and inflection points of f(x)=lnx/sqrt(x)?

Feb 17, 2017

graph{lnx/sqrtx [-0,5, 1000, -2.88, 2]}

#### Explanation:

We start by observing that the function:

$f \left(x\right) = \ln \frac{x}{\sqrt{x}}$

is defined in the interval: $x \in \left(0 , + \infty\right)$, that it is negative for $x < 1$, positive for $x > 1$ and has a single zero for $x = 1$

We can analyze the behavior at the limits of the domain:

${\lim}_{x \to {0}^{+}} \ln \frac{x}{\sqrt{x}} = - \infty$

${\lim}_{x \to \infty} \ln \frac{x}{\sqrt{x}} = 0$

so the function has the line$x = 0$ for vertical asymptote and the line $y = 0$ has horizontal asymptote.

Let's calculate the first and second derivative of the function:

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

$f ' ' \left(x\right) = \frac{\left(- \frac{2 x \sqrt{x}}{x}\right) - \left(\left(2 - \ln x\right) 3 \sqrt{x}\right)}{4 {x}^{3}} = \frac{- 2 \sqrt{x} - 6 \sqrt{x} + 3 \ln x \sqrt{x}}{4 {x}^{3}} = \frac{3 \ln x - 8}{4 {x}^{2} \sqrt{x}}$

Now we can see that:

(1) $f ' \left(x\right) > 0$ for $2 - \ln x > 0$ that is for $x < {e}^{2}$

so the function is increasing in the interval $\left(0 , {e}^{2}\right)$ and decreasing in $\left({e}^{2} , + \infty\right)$ and has a single critical value in $x = {e}^{2}$ that is a local maximum.

(2) $f ' ' \left(x\right) > 0$ for $3 \ln x - 8 > 0$ or $x > {e}^{\frac{8}{3}}$

so the function is concave down in the interval $\left(0 , {e}^{\frac{8}{3}}\right)$ and concave up in $\left({e}^{\frac{8}{3}} , + \infty\right)$ and has a single inflection point in $x = {e}^{\frac{8}{3}}$.