Q. 79 - Calculus

Question

Sketch careful, labeled graphs of each function $f$ in Exercises $57–82$ by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f$ and $f\text{'}$ , and examine any relevant limits so that you can describe all key points and behaviors of $f$.

 $f\left(x\right)=\frac{\ln x}{x}$

Step-by-Step Solution

Answer

The  $f$ is defined everywhere, and the root of the $f$ is at $x=0$, The function is positive everywhere except on $x=0$. The $f$ have a local minimum at $x=-3$, and local maximum at $x=0$. The $f$ is increasing on $(-3,\infty )$ and negative elsewhere. $\underset{x\to -\infty }{\lim }f\left(x\right)=0$. So, there is horizontal asymptote on the left at$y=0$. $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$; so there is no horizontal asymptote on the right.

Step 1. Given data

We have the given function,$f\left(x\right)=\frac{\ln x}{x}$

that is,$y=\frac{\ln x}{x}$

Step 2. Graph

The point table of the function is given by,

x	y	(x,y)
1	0	(1,0)
2	0.3465	(2,0.3465)
3	0.366	(3,0.366)
4	0.3465	(4,0.3465)

The graph of the function is,

Step 3. Critical points

To find the critical points, let$f^{\text{'}}\left(x\right)=0$

$$\begin{gathered} \begin{array}{rl}\frac{d}{dx}\left(\frac{\ln x}{x}\right)& =0\\ \frac{x\cdot \frac{1}{x}-\ln x\cdot 1}{x^2}& =0\\ \frac{1-\ln x}{x^2}& =0\\ 1-\ln x& =0\end{array} \\ \begin{array}{rl} \ln x& =1\\ x& =e\end{array} \end{gathered}$$

Hence, $f$ has a critical points are at $x=e$.

Local minima at$x=-3$, and local maxima at $x=0$.

Step 4. Sign chart

The sign chart of the function is,

Step 5. Roots of the function

For the roots of the functions be,

$\begin{array}{r}{x}^3{e}^x=0\\ x^3=0\\ x=0\end{array}$

Hence the function has a root $x=0$

Again,

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to \mathrm{\infty }}{lim}{x}^3{e}^x\\ & =\mathrm{\infty }\\ \underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to -\mathrm{\infty }}{lim}{x}^3{e}^x\\ & =0\end{array}$

Hence, the function $f$ is defined everywhere and the root of the function is at $x=0$, The function is positive everywhere except on $x=0$. The function have a local minimum at $x=-3$; and local maximum at $x=0$. The function is increasing on $(-3,\infty )$ and negative elsewhere. $\underset{x\to -\infty }{\lim }f\left(x\right)=0$. Therefore, there is a horizontal asymptote on the left at $y=0$. $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$, so there is no horizontal asymptote on the right.