Chapter 2.6, Problem 55E

Solution

The intercepts, analyse and graph the function.

The intercept are $\left(1.26,0\right)$ and $\left(0,1\right)$ .

Given:

The function is $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ .

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 $f\left(x\right)/g\left(x\right)$ . Hence, no horizontal asymptotes.

Calculation:

Consider the function, $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ .

Find the intercept, the x -intercept is given by zeros of numerator that are not zero of denominator.

So, find the zeros of the numerator as,

  $\begin{array}{c}{x}^3-2x^2+x-1=0\\ x=1.75\end{array}$

Thus, $x=1.75$ . The x- intercepts is $\left(1.75,0\right)$ .

And y -intercept is given by $f\left(0\right)$ , so from $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ .

  $\begin{array}{c}f\left(x\right)=\frac{{\left(0\right)}^3-2{\left(0\right)}^2+\left(0\right)-1}{2\left(0\right)-1}\\ =1\end{array}$

Thus y -intercept is $\left(0,1\right)$ .

Hence the intercept are $\left(1.26,0\right)$ and $\left(0,1\right)$ .

Find vertical asymptotes, that is given by zeros of denominator $2x-1$ ,

Since, the zeros of denominator of $f\left(x\right)$ is $x=-1/2$ , Vertical asymptotes is line $x=-1/2$ .

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ 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 the quotient of $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ . Rewrite $f\left(x\right)$ using division as follows,

  

  $\begin{array}{l}-\underset{¯}{\left(-\frac{3}{2}{x}^2+\frac{3}{4}x\right)}\\ \frac{1}{4}x-1\\ \underset{¯}{-\left(\frac{1}{4}x-\frac{1}{8}\right)}\\ -\frac{7}{8}\end{array}$

Thus, write $f\left(x\right)$ as,

  $f\left(x\right)=\left(\frac{1}{2}{x}^2-\frac{3}{4}x-\frac{1}{8}\right)-\frac{7}{8\left(2x-1\right)}$

Hence, end behaviour asymptotes is $y=\frac{1}{2}{x}^2-\frac{3}{4}x-\frac{1}{8}$ .

To graph $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ , compute the value of $f\left(x\right)$ for chosen values of x as follows:

  

The graph of $f\left(x\right)=\frac{x^3-2x^2+x-1}{2x-1}$ obtain is as:

  

Interpretations form graph:

1) Domain of $f\left(x\right)$ is $\left(-\infty ,0.5\right)\cup \left(0.5,\infty \right)$

2) Range is $\left(-\infty ,\infty \right)$ .

3) Continuous everywhere except $x=\frac{1}{2}$

4) Increasing in $\left(-0.18,0.5\right)\cup \left(0.5,\infty \right)$ and decreasing in $\left(-\infty ,-0.18\right)$ .

5) No Local maxima and local minima at $\left(-0.18,0.91\right)$ .

6) Not symmetric.

7) Unbounded.

Conclusion:

The intercept are $\left(1.26,0\right)$ and $\left(0,1\right)$ .