Chapter 2.6, Problem 80E

Solution

To Sketch: The given function and evaluate it as piecewise-defined function.

Find the graph's asymptotes and calculate the $x$ and $y$ intercepts.

Given function: $f\left(x\right)=\frac{2-2x}{\left|x\right|+1}$

Graph:

  

Piecewise function:

The given function as a piecewise function as:

  $f\left(x\right)=\left\{\begin{array}{ll}\frac{2-2x}{x+1}\hfill & x\ge 0\hfill \\ \frac{2-2x}{-x+1}\hfill & x<0\hfill \end{array}\right.$

Find the asymptotes:

Check for vertical asymptote:

If $x\ge 0$ there should be a vertical asymptote if

  $\begin{array}{c}x+1=0\\ x=-1\end{array}$

but as $x$ is not negative, it is not possible.

Thus, the function$f\left(x\right)=\frac{2-2x}{\left|x\right|+1}$has no vertical asymptotes.

Check for horizontal asymptote:

The asymptote is the ratio between the leading coefficients, where numerator and denominator have the same degree.

For $x\ge 0$ , there is a horizontal asymptote at,

  $\begin{array}{l}y=\frac{-2}{1}\\ y=-2\end{array}$

For $x<0$ there is a horizontal asymptote at,

  $\begin{array}{l}y=\frac{-2}{-1}\\ y=2\end{array}$

In other words, here in piece function $f\left(x\right)=\left\{\begin{array}{ll}\frac{2-2x}{x+1}\hfill & x\ge 0\hfill \\ \frac{2-2x}{-x+1}\hfill & x<0\hfill \end{array}\right.$ both the degree of numerator and denominator is equal to $1$ so, the end behaviour asymptote is horizontal asymptote:

  $y=\pm 2$

Find the $x-$intercepts:

The $x-$ intercept occur at the zero of the numerators:

  $\begin{array}{c}2-2x=0\\ 2x=2\\ x=1\end{array}$

Thus the $x-$ intercepts of the function are $\left(1,0\right)$ .

Find the $y-$intercepts:

To find the $y-$ intercepts, substitute $x=0$ in the function $f\left(x\right)=\frac{2-2x}{\left|x\right|+1}$ :

  $\begin{array}{l}f\left(0\right)=\frac{2-2\left(0\right)}{\left|0\right|+1}\\ f\left(0\right)=\frac{2-0}{0+1}\\ f\left(0\right)=\frac{2}{1}\\ f\left(0\right)=2\end{array}$

Thus the $y-$ intercepts of the function are $\left(0,2\right)$ .