Chapter 4.3, Problem 15E

(a)

To determine

To find:The interval on which $f$ is increasing.

Explanation of Solution

Given: The function is $f\left(x\right)=\frac{\ln x}{\sqrt{x}}$

Find the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{\frac{1}{\sqrt{x}}-\frac{\ln x}{2\sqrt{x}}}{x}\\ =\frac{-\ln x+2}{2x^{3/2}}\end{array}$

On solving further,

  $x=e^2$

The function increases in the interval $\left(0,e^2\right)$ .

The function decreases in the interval $\left(e^2,\infty \right)$

(b)

To determine

To find :The local maxima and minima.

Explanation of Solution

The function increases in the interval $\left(0,e^2\right)$ .

The function decreases in the interval $\left(e^2,\infty \right)$

The local maximaat $\left(e^2,f\left(e^2\right)\right)=\left(e^2,\frac{2}{e}\right)$ .

(c)

To determine

To find :The interval ofconcavity and points of inflection

Explanation of Solution

The graph is concave up when $f^{″}\left(x\right)>0$ and concave downward when $f^{″}\left(x\right)<0$ .

The derivative of $f^{\prime }\left(x\right)$ is positive when $f^{\prime }\left(x\right)$ is increasing.

The derivative of $f^{\prime }\left(x\right)$ is negative when $f^{\prime }\left(x\right)$ is decreasing

Calculate the second derivative.

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{3\left(\ln \left(x\right)-2\right)}{4x^{5/2}}-\frac{1}{2x^{5/2}}\\ =\frac{3\ln x-8}{4x^{5/2}}\\ 8=3\ln x\\ x=e^{8/3}\end{array}$

The function is concave up for $\left(e^{8/3},\infty \right)$ .

The function is concave down for $\left(0,e^{8/3}\right)$ .

The point of inflection is $e^{8/3}$ .