Chapter 4, Problem 15CR

Given the function $f\left(x\right)=\frac{x+5}{x-1}$

What is the domain of $f$?

Is the point $\left(2,6\right)$ on the graph of $f$?

If $x=3$, what is $f\left(x\right)$? What point is on the graph of $f$?

If $f\left(x\right)=9$, what is x? What point is on the graph of $f$?

Is $f$ a polynomial or a rational function?

To determine

The domain of the function $f\left(x\right)=\frac{x+5}{x-1}$

Answer to Problem 15CR

Solution:

The domain of the function $f\left(x\right)=\frac{x+5}{x-1}$ is $\left(-\infty ,1\right)\cup \left(1,\infty \right)$

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x+5}{x-1}$

Explanation:

Domain of a function is a set of values of $x$, where the function is defined

Since $f\left(x\right)$ is a rational function, it is defined except for the zeros of the denominator.

Set the denominator to zero and simplify.

By simplifying $x-1=0$,

$x=1$

Thus, the function $f\left(x\right)$ is defined on a set of real numbers, except for $x=1$

Hence, the domain of the function $f\left(x\right)=\frac{x+5}{x-1}$ is $\left(-\infty ,1\right)\cup \left(1,\infty \right)$

To determine

Whether the point $\left(2,6\right)$ is on the graph of $f\left(x\right)=\frac{x+5}{x-1}$

Answer to Problem 15CR

Solution:

The point $\left(2,6\right)$ is not on the graph of $f\left(x\right)=\frac{x+5}{x-1}$

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x+5}{x-1}$

Explanation:

By comparing the given point $\left(2,6\right)$ with $\left(x,y\right)$

$x=2$ and $y=f\left(2\right)=6$, it means the value of the function corresponding to $x=2$ is $6$

By substituting $x=2$ in the function $f\left(x\right)=\frac{x+5}{x-1}$

$f\left(2\right)=\frac{2+5}{2-1}$

$=\frac{7}{1}$

$=7$

$\Rightarrow f\left(2\right)=7$

Thus, $f\left(2\right)=7\ne 6$

The value of the function corresponding to $x=2$ is not equal to $6$

Hence, the point $\left(2,6\right)$ is not on the graph of $f\left(x\right)=\frac{x+5}{x-1}$

To determine

The value of $f\left(x\right)$, when $x=3$ and determine what point is on the graph of $f$, where $f\left(x\right)=\frac{x+5}{x-1}$.

Answer to Problem 15CR

Solution:

The value of $f\left(x\right)$, when $x=3$ is $4$ and $\left(3,4\right)$ is a point on the graph of $f$

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x+5}{x-1}$

Explanation:

Substitute $x=3$ in $f\left(x\right)=\frac{x+5}{x-1}$

$f\left(3\right)=\frac{3+5}{3-1}$

$=\frac{8}{2}$

$=4$

$\Rightarrow f\left(3\right)=4$

Thus, the value of $f\left(x\right)$, when $x=3$ is $4$

For $x=3$, the corresponding value of function is $f\left(3\right)=4$

Hence, $\left(3,4\right)$ is a point on the graph of $f\left(x\right)=\frac{x+5}{x-1}$

To determine

The value of $x$, when $f\left(x\right)=9$ and determine what point is on the graph of $f$, where $f\left(x\right)=\frac{x+5}{x-1}$.

Answer to Problem 15CR

Solution:

The value of $x$ is $1.75$, when $f\left(x\right)=9$ and $\left(1.75,9\right)$ is a point on the graph of $f$.

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x+5}{x-1}$

Explanation:

Substitute $f\left(x\right)=9$ in $f\left(x\right)=\frac{x+5}{x-1}$

$9=\frac{x+5}{x-1}$

$\Rightarrow 9\left(x-1\right)=x+5$

$\Rightarrow 9x-9=x+5$

$\Rightarrow 9x-x=9+5$

$\Rightarrow 8x=14$

$\Rightarrow x=\frac{14}{8}=\frac{7}{4}=1.75$

Thus, the value of $x$ is $1.75$, when $f\left(x\right)=9$

For $f\left(x\right)=9$, the corresponding value of $x$ is $1.75$

Hence, $\left(1.75,9\right)$ is a point on the graph of $f\left(x\right)=\frac{x+5}{x-1}$

To determine

Whether the function $f\left(x\right)=\frac{x+5}{x-1}$ is a polynomial function or a rational function.

Answer to Problem 15CR

Solution:

The function $f\left(x\right)$ is a rational function.

Explanation of Solution

Given information:

The function $f\left(x\right)=\frac{x+5}{x-1}$

Explanation:

Since, the function $f\left(x\right)=\frac{x+5}{x-1}$ is in the form of $\frac{P}{Q}$, where $P$ and $Q$ are polynomial functions, hence the function $f\left(x\right)$ is a rational function.