Chapter 3, Problem 66RE

(a)

To determine

To find: The interval on which the function is increasing.

Explanation of Solution

Given: The given functionis $f\left(x\right)=\ln x/x$ .

Consider the function.

  $f\left(x\right)=\ln x/x$

Differentiate the above equation.

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

  $f^{\prime }\left(x\right)=0$ when $\ln \left(x\right)=1$ which is when $x=e$ . Thus, $f$ switches from being increasing to decreasing or vice versaat $x=e$ .

If $x=1$ hen $f^{\prime }\left(x\right)=1>0$ and if $x=e^2$ then $f^{\prime }\left(x\right)=\frac{-1}{e^4}<0$ . Thus, $f^{\prime }\left(x\right)$ is increasing in the interval $\left(-\infty ,e\right)$ .

Therefore, the interval in which the function is increasing is $\left(-\infty ,e\right)$ .

(b)

To determine

To find : The interval on which the function is concave upward.

Explanation of Solution

Given: The given function is $f\left(x\right)=\ln x/x$ .

Consider the function.

  $f\left(x\right)=\ln x/x$

Differentiate the above equation.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{x\left(\frac{1}{x}\right)-\ln \left(x\right)\cdot 1}{x^2}\\ =\frac{1-\ln x}{x^2}\\ f^{″}\left(x\right)=\frac{-3x+2x\ln x}{x^4}\end{array}$

  $f^{″}\left(x\right)=0$ when $2x\ln \left(x\right)=3x$ which is when $x=e^{3/2}$ . Thus,

  $f$ switches from being concave up to concave down or vice versa at $x=e^{3/2}$ .

If $x=1$ then $f^{″}\left(x\right)=-3<0$ and if $x=e^2$ then $f^{″}\left(x\right)=\frac{1}{e^6}>0$ . Thus, $f\left(x\right)$ is concave up in the interval $\left(e^{3/2},\infty \right)$ .

Therefore, the function $f\left(x\right)$ is concave up in the interval $\left(e^{3/2},\infty \right)$ .