Question

In Problems 77 – 82, determine where each rational function is undefined. Determine whether an asymptote or a hole appears at such numbers.

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

Short answer

The function is not defined at $x=-\sqrt[3]{2}$and $x=-1$.

At $x=-\sqrt[3]{2}$the function has an asymptote and at $x=-1$a hole appears in the graph of the function.

Step-by-step solution

Step 1. Given Information

We are given a rational function $R\left(x\right)=\frac{x^3+2x^2+x}{x^4+x^3+2x+2}$.

We need to find the values where the function is not defined and to determine whether there is an asymptote or a hole appears at those points.

Step 2. Find the points where the function is undefined  

A rational function is not defined at those values of xwhere the denominator is zero. So equate denominator function equal to zero and find the zeros as

$\begin{array}{rcl}{x}^4+x^3+2x+2& =& 0\\ x^3(x+1)+2(x+1)& =& 0\\ (x+1)(x^3+2)& =& 0\end{array}$

So

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

and

$\begin{array}{rcl}{x}^3+2& =& 0\\ x^3& =& -2\\ x& =& -\sqrt[3]{2}\end{array}$

Thus at $x=-1$and $\begin{array}{rcl}x& =& -\sqrt[3]{2}\end{array}$the denominator is zero and so the rational function is not defined at these values.

Step 3. Find the limit at  x=-1

The limit of the function at $x=-1$is given as

$\begin{array}{rcl}\underset{x\to -1}{\lim }\frac{x^3+2x^2+x}{x^4+x^3+2x+2}& =& \underset{x\to -1}{\lim }\frac{x{(x+1)}^2}{(x+1)(x^3+2)}\\ & =& \underset{x\to -1}{\lim }\frac{x(x+1)}{x^3+2}\\ & =& 0\end{array}$

So the limit of $R\left(x\right)$at $x=-1$is $0$but the function is not defined at $x=-1$. This means that there is a hole and discontinuity at the point $(-1,0)$.

Step 4. Find the limit at x=-23

The limit at $x=-\sqrt[3]{2}$is given as

$\begin{array}{rcl}\underset{x\to -\sqrt[3]{2}}{\lim }\frac{x^3+2x^2+x}{x^4+x^3+2x+2}& =& \underset{x\to -\sqrt[3]{2}}{\lim }\frac{x{(x+1)}^2}{(x+1)(x^3+2)}\\ & =& \underset{x\to -\sqrt[3]{2}}{\lim }\frac{x(x+1)}{x^3+2}\end{array}$

So it can be seen that as $x=-\sqrt[3]{2}$, the function value approaches infinity. From the left hand side the graph approaches negative infinity and from the right hand side the graph approaches positive infinity.

So $R\left(x\right)$is discontinuous at $x=-\sqrt[3]{2}$and there is a vertical asymptote at $x=-\sqrt[3]{2}$.