Chapter 2.6, Problem 48E

Solution

The end behaviour asymptotes of function f and graph it in two window a) One showing around vertical asymptote(s). b) One showing a graph of f that resembles the end behaviour asymptote.

The end behaviour asymptotes of $f\left(x\right)=\frac{x^3+1}{x-1}$ is $y=\left(x^2+x+1\right)$ , with a vertical asymptote is $x=1$ .

Given:

The function is $f\left(x\right)=\frac{x^3+1}{x-1}$ .

Concept Used:

The x -intercept is given by zeros of numerator that are not zero of denominator. And y -intercept is given by $f\left(0\right)$ .

And,

If a polynomial function in the form $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ , then, vertical asymptotes is given by real zeros of denominator provided it should not be zero of numerator.

And,

The end behaviour asymptote given by $\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\text{ or }\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is horizontal asymptotes.

The condition can be concluded as,

1) If $n<m$ , the end behaviour is horizontal asymptotes that is $y=0$ ,

2) If $n=m$ the end behaviour is horizontal asymptotes is horizontal asymptotes that is $y=\frac{a_n}{b_n}$

3) If $n>m$ , the quotient $q\left(x\right)$ of $f\left(x\right)/g\left(x\right)$ is the end behaviour asymptotes, where $q\left(x\right)$ is given by $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ with $r\left(x\right)$ is remainder of $f\left(x\right)/g\left(x\right)$ .

Calculation:

Consider the function, $f\left(x\right)=\frac{x^3+1}{x-1}$ .

Find vertical asymptotes, so find the zeros of denominator,

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

Since, the zero of denominator is vertical asymptotes, so vertical asymptote is $x=1$ .

Therefore, $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=\infty \text{ or }-\infty$

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{x^3+1}{x-1}$ with $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ .

Since, $n>m$ end behaviour asymptotes is given by quotient of $f\left(x\right)=\frac{x^3+1}{x-1}$ . So rewrite the function using long division as follows,

  

Thus, $f\left(x\right)=\frac{x^3+1}{x-1}$ can be rewritten as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3+1}{x-1}\\ =\frac{\left(x-1\right)\left(x^2+x+1\right)+2}{x-1}\\ =\left(x^2+x+1\right)+\frac{2}{x-1}\end{array}$

Or, $f\left(x\right)=\left(x^2+x+1\right)+\frac{2}{x-1}$ .

Hence, the end behaviour asymptotes is, $y=\left(x^2+x+1\right)$ .

Now, graph the end behaviour asymptote together with f in two window, a) One showing around vertical asymptote and b) One showing a graph of f that resembles the end behaviour asymptote.

a) The graph of f with end behaviour asymptote, $y=\left(x^2+x+1\right)$ around vertical asymptotes, $x=1$ is as follows,

  

And, b) The graph of f that resembles the end behaviour asymptote is as follows,

  

Clearly, the graph of f resembles the end behaviour asymptote.

Conclusion:

The end behaviour asymptotes of $f\left(x\right)=\frac{x^3+1}{x-1}$ is $y=\left(x^2+x+1\right)$ , with a vertical asymptote is $x=1$ .