Chapter 3.7, Problem 29E

  $\left(a\right)$

To determine

To calculate: For the function, $f\left(x\right)=x\ln x$ , the intervals in which the function is decreasing.

Answer to Problem 29E

The function $f\left(x\right)=x\ln x$ is decreasing in the interval $\left(0,\frac{1}{e}\right)$ .

Explanation of Solution

Given information:

The function, $f\left(x\right)=x\ln x$ .

Formula used:

Let a function g be continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every xin $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every xin $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function, $f\left(x\right)=x\ln x$ .

Differentiate the function with respect to x . Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Apply it,

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{d}{dx}\left(x\ln x\right)\\ =x\frac{d}{dx}\left(\ln x\right)+\left(\ln x\right)\frac{d}{dx}x\\ =x\cdot \frac{1}{x}+\ln x\left(1\right)\\ =1+\ln x\end{array}$

Recall if a function g is continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every xin $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

Apply it,

  $\begin{array}{c}1+\ln x<0\\ \ln x<-1\\ e^{\ln x}<e^{-1}\\ x<\frac{1}{e}\end{array}$

So, $f\text{'}\left(x\right)<0$ , the function $f\left(x\right)=x\ln x$ is decreasing in interval $\left(0,\frac{1}{e}\right)$ .

Thus, the function $f\left(x\right)=x\ln x$ is decreasing in the interval $\left(0,\frac{1}{e}\right)$ .

  $\left(b\right)$

To determine

To calculate: For the function, $f\left(x\right)=x\ln x$ , the interval in which the function is concave upward.

Answer to Problem 29E

The function $f\left(x\right)=x\ln x$ is concave upward in the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given information:

The function, $f\left(x\right)=x\ln x$ .

Formula used:

Let a function g be continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every xin $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every xin $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If the second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every xin $\left(c,d\right)$ , then the function g is concave upward on interval $\left[c,d\right]$ .

Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Calculation:

Consider the function, $f\left(x\right)=x\ln x$ .

Differentiate the function with respect to x . Product rule for differentiation is $\frac{d}{dx}\left(f\cdot g\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)$ where f and g are functions of x .

Apply it,

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{d}{dx}\left(x\ln x\right)\\ =x\frac{d}{dx}\left(\ln x\right)+\left(\ln x\right)\frac{d}{dx}x\\ =x\cdot \frac{1}{x}+\ln x\left(1\right)\\ =1+\ln x\end{array}$

Differentiate again with respect to x .

  $\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(1+\ln x\right)\\ =\frac{d}{dx}\left(\ln x\right)+\frac{d}{dx}\left(1\right)\\ =\frac{1}{x}\end{array}$

Recall if a function g is continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If the second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every xin $\left(c,d\right)$ , then the function g is concave upward on interval $\left[c,d\right]$ .

Apply it, so $\frac{1}{x}>0$ this implies $x>0$ .

So, $f\text{'}\text{'}\left(x\right)>0$ , the function $f\left(x\right)=x\ln x$ is concave upwardin interval $\left(0,\infty \right)$ .

Thus, the function $f\left(x\right)=x\ln x$ is concave upward in the interval $\left(0,\infty \right)$ .