Chapter 6.R, Problem 81E

To determine

To Draw:

Graph of the function $y=x\ln x$.

Solution:

The graph is sketched below.

Given:

$y=x\ln x$

Explanation:

Consider, the function

$y=x\ln x$

(1) Domain of the function y:

Domain of $\ln x\text{ is }\left\{x\in ℝ\text{: }x>0\right\}$. Therefore,

Domain of $x\ln x\text{ is }\left\{x\in ℝ\text{: }x>0\right\}.$

Thus, domain of $y(x)$ is $\left\{x\in ℝ\text{: }x>0\right\}$

(2) Intercepts:

There is no y-intercept as x = 0 is not in domain of y.

For x-intercept, put

 $\begin{array}{c}0=y(x)\\ =x\ln x\end{array}$

But x > 0, therefore

$\begin{array}{l}\ln x=0\\ \text{This implies that }x=e^0=1.\text{ Thus, }x\text{-intercept is }(1,0).\end{array}$

(3)

$\begin{array}{c}y(-x)=-x\ln (-x)\\ \ne x\ln x\\ \ne -x\ln x\end{array}$

This implies that y is neither even nor odd function.

(4) Asymptotes:

Since $\ln x$ has vertical asymptote at x =0, therefore, the possible vertical asymptote for y can be exist at x =0.  Consider

$\begin{array}{c}\underset{x\to 0^{+}}{\lim }y(x)=\underset{x\to 0^{+}}{\lim }x\ln (x)\\ =\underset{x\to 0^{+}}{\lim }\frac{x}{\ln x}\left(\frac{0}{0}\text{ form }\right)\\ =\underset{x\to 0^{+}}{\lim }\frac{\left(x\right)\text{'}}{\left(\ln x\right)\text{'}}\text{ (By l'Hospital' rule)}\\ \text{=}\underset{x\to 0^{+}}{\lim }\frac{1}{\left(1/x\right)}\\ =0\end{array}$

 This shows that at x =0, y has no vertical asymptote.

Horizontal asymptotes:

$\begin{array}{c}\underset{x\to \infty }{\lim }y(x)=\underset{x\to \infty }{\lim }x\ln (x)\\ =\infty \end{array}$

Thus, y has no horizontal asymptotes..

(5) Increasing or decreasing:

Consider

$y=x\ln x$

Differentiate it with respect to x, we get

$\begin{array}{c}y\text{'}=\frac{d}{dx}x\ln x\\ =\frac{d}{dx}(x)\ln x+x\frac{d}{dx}\ln x\left(\text{By product rule}\right)\\ =1\times \ln x+x\times \frac{1}{x}\\ \text{=}\ln x+1\end{array}$

Put $y\text{'}(x)=0$. This implies that

$\begin{array}{l}\ln x+1=0\\ \text{So, }\\ \ln x=-1\\ \text{This gives that}\\ x=e^{-1}\end{array}$

Consider

Interval	Point in the interval	Value of $y\text{'}(x)=\ln x+1$	Sign of $y\text{'}$	Increasing or decreasing
$(0,e^{-1})$	$e^{-2}$	$\begin{array}{l}\ln e^{-2}+1=-2+1\\ =-1\end{array}$	$-\text{ve}$	Decreasing
$(e^{-1},\infty )$	$e$	$\begin{array}{l}\ln e+1=1+1\\ =2\end{array}$	$\text{+ve}$	Increasing

This implies that $y$ is decreasing in the interval $(0,e^{-1})$ and increasing in $(e^{-1},\infty )$.

(6) Local maxima or minima:

Consider

$y\text{'}=\ln x+1$

Differentiate it with respect to x, we get

$\begin{array}{c}y\text{'}\text{'}=\frac{d}{dx}\left(\ln x+1\right)\\ =\frac{1}{x}\text{+0 }\\ \text{=}\frac{1}{x}\end{array}$

At $x=e^{-1}$

$y\text{'}\text{'}\left(e^{-1}\right)=\frac{1}{e^{-1}}>0$

Therefore, $x=e^{-1}$ is point of minima.

(7) Inflection point:

Since $y\text{'}\text{'}(x)\text{=}\frac{1}{x}>0\text{ for all domain point, therefore, }y\text{ has no inflection point}\text{.}$

Thus, by the above information’s graph of y is: