Q. 67 - 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 $f$.

$f\left(x\right)=\frac{x+1}{x-1}$.

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=\frac{x+1}{x-1}$ is,

Step 1 . Given information

$f\left(x\right)=\frac{x+1}{x-1}$.

Step 2 . Let y = x + 1 x - 1 .

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-2$	$\frac{1}{3}$	$\left(-2,\frac{1}{3}\right)$
$-1$	$0$	$\left(-1,0\right)$
$0$	$-1$	$\left(0,-1\right)$
$2$	$3$	$\left(2,3\right)$
$4$	$\frac{5}{3}$	$\left(4,\frac{5}{3}\right)$

Step 3 . The graph for the function is,

Step 4 . Now for critical point f ' x = 0 .

$$\begin{gathered} \frac{d}{dx}\left(\frac{x+1}{x-1}\right)=0 \\ \frac{(x-1)·1-(x+1)·1}{(x-1{)}^2}=0 \\ \frac{x-1-x-1}{(x-1{)}^2}=0 \\ \frac{-2}{(x-1{)}^2}=0 \end{gathered}$$

Therefore, $f$ does not have a critical point. So, it has no local extremum.

Step 5 . The sign chart of f is shown below:

For roots of the function,

$$\begin{gathered} \frac{x+1}{x-1}=0 \\ x+1=0 \\ x=-1 \end{gathered}$$

Step 6 . Therefore, the function f is increasing on - ∞ , 0 and decreasing on 0 , ∞ .

Again,

$\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$

                   $=$ Doesn't exist.

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right) \\ =\infty \end{gathered}$$

Therefore, the function is defined on $\left(0,\infty \right)$ and the root at $x=-1$. Positive on $(-\infty ,-1)\cup (1,\infty )$ and negative at $\left(-1,1\right)$ except at $x=0$.