Chapter 14.3, Problem 78AYU

$R\left(x\right)=\frac{x^3+x^2+3x+3}{x^4+x^3+2x+2}$

To determine

To find: The numbers at which $R$ is undefined and asymptotes or holes for $R$.

Answer to Problem 78AYU

Undefined for $x=1$ and $x=-\sqrt[3]{2}$.

Vertical asymptote: $x+1=0$; $x=-\sqrt[3]{2}$.

Horizontal asymptote: $y=0$.

Explanation of Solution

Given:

$R\left(x\right)=\frac{x^3+x^2+3x+3}{x^4+x^3+2x+2}$

Calculation:

$R\left(x\right)=\frac{x^3+x^2+3x+3}{x^4+x^3+2x+2}$

$R\left(x\right)=\frac{\left(x^2+3\right)\left(x+1\right)}{\left(x^3+2\right)\left(x+1\right)}$

The denominator of $R\left(x\right)$ becomes zero when

$\left(x^3+2\right)\left(x+1\right)=0$

$x=-\sqrt[3]{2}$; $x=-1$

Thus $R$ is undefined for $x=-1$ and $x=-\sqrt[3]{2}$.

This function has vertical asymptotes $x+1=0$ and $x=-\sqrt[3]{2}$ and horizontal asymptote $y=0$.