Chapter 4, Problem 1RE

Finding the Domain of a Rational Function In Exercises 1-4, Find the domain of the function and discuss the behavior of $f$ near any excluded $x\text{-values}\text{.}$

$f(x)=\frac{3x}{x+10}$

To determine

To calculate: The domain of the function $f\left(x\right)=\frac{3x}{x+10}$. Also, discuss the behavior of the function near any excluded $x-\text{values}$.

Answer to Problem 1RE

The domain of $f$ is all real numbers except $x=-10$. When $x$ approaches $-10$ from the left side, $f\left(x\right)$ will increase without bound. As $x$ approaches $-10$ from the right side, $f\left(x\right)$ will decrease without bound.

Explanation of Solution

Calculation:

Consider the function $f\left(x\right)=\frac{3x}{x+10}$.

To find the domain, make the denominator zero and get the value of $x$. Then, the domain includes all the real numbers except $x-\text{values}$.

As the denominator becomes zero at $x=-10$, the domain of the function $f$ is all real numbers except $x=-10$.

Now, the behavior of $f$ can be determined from a table by evaluating $f\left(x\right)$ to the left and right of $x=-10$.

First construct the table when $x$ approaches from the left side of $x=-10$.

For $x=-11$, the value of $f\left(x\right)$ will be,

$\begin{array}{c}f\left(x\right)=\frac{3x}{x+10}\\ =\frac{3\left(-11\right)}{-11+10}\\ =33\end{array}$

For $x=-10.5$, the value of $f\left(x\right)$ will be,

$\begin{array}{c}f\left(x\right)=\frac{3x}{x+10}\\ =\frac{3\left(-10.5\right)}{-10.5+10}\\ =63\end{array}$

Similarly, the values of $f\left(x\right)$ for corresponding values of $x$ are in the following table:

$\begin{array}{|cccccc|}\hline x& -11& -10.5& -10.1& -10.01& -10.001\\ f\left(x\right)& 33& 63& 303& 3003& 30003\\ \hline\end{array}$

Now, construct the table when $x$ approaches from the right side of $x=-10$.

For $x=-9.999$, the value of $f\left(x\right)$ will be,

$\begin{array}{c}f\left(x\right)=\frac{3\left(-9.999\right)}{-9.999+10}\\ =-29997\end{array}$

Also, for $x=-9.99$, the value of $f\left(x\right)$ will be,

$\begin{array}{c}f\left(x\right)=\frac{3\left(-9.99\right)}{-9.99+10}\\ =-2997\end{array}$

Similarly, the values of $f\left(x\right)$ for corresponding values of $x$ are in the following table:

$\begin{array}{|cccccc|}\hline x& -9.999& -9.99& -9.9& -9.5& -9\\ f\left(x\right)& -29997& -2997& -297& -57& -27\\ \hline\end{array}$

Therefore, from the table, it is observed that as $x$ approaches $-10$ from the left side, $f\left(x\right)$ will increase without bound. As $x$ approaches $-10$ from the right side, $f\left(x\right)$ will decrease without any bound.