Q. 14 - Precalculus Enhanced with Graphing Utilities

Question

Find $f+g,f-g,f\cdot g$ and $\frac{f}{g}$ for each pair of functions. State the domain of each of these functions.

$$\begin{gathered} f\left(x\right)=\frac{x+1}{x-1} \\ g\left(x\right)=\frac{1}{x} \end{gathered}$$

Step-by-Step Solution

Answer

For  $f\left(x\right)=\frac{x+1}{x-1}; g\left(x\right)=\frac{1}{x}$, we get:

$f+g=\frac{x^2+x+x+1}{x^2-1}$ and its domain is $\left\{x\right|x\ne 0,1\}$.

$f-g=\frac{x^2+1}{x(x-1)}$ and its domain is $\left\{x\right|x\ne 0,1\}$.

$f\cdot g=\frac{x+1}{x(x-1)}$ and its domain is $\left\{x\right|x\ne 0,1\}$.

$\frac{f}{g}=\frac{x(x+1)}{x-1}$ and ints domain is $\left\{x\right|x\ne 1\}$.

Step 1. Given information

We have been given:

$$\begin{gathered} f\left(x\right)=\frac{x+1}{x-1} \\ g\left(x\right)=\frac{1}{x} \end{gathered}$$

We have to find $f+g,f-g,f\cdot g$ and $\frac{f}{g}$ for the pair of functions and state the domain of each of these functions.

Step 2. Find f + g and its domain.

We know $f+g=f\left(x\right)+g\left(x\right)$.

Therefore,

$$\begin{gathered} f+g \\ =\frac{x+1}{x-1}+\frac{1}{x} \\ =\frac{x(x+1)+(x-1)}{x(x-1)} \\ =\frac{x^2+x+x+1}{x(x-1)} \end{gathered}$$

The domain will be the set of all real numbers except those that make the denominator zero.

$$\begin{gathered} x(x-1)=0 \\ x=0,1 \end{gathered}$$

Therefore, the domain will be $\left\{x\right|x\ne 0,1\}$.

Step 3. Find f - g and its domain.

We know $f-g=f\left(x\right)-g\left(x\right)$.

Therefore,

$$\begin{gathered} f-g \\ =\frac{x+1}{x-1}-\frac{1}{x} \\ =\frac{x(x+1)-1(x-1)}{x(x-1)} \\ =\frac{x^2+x-x+1}{x(x-1)} \\ =\frac{x^2+1}{x(x-1)} \end{gathered}$$

The domain will be the set of all real numbers except those that make the denominator zero.

$$\begin{gathered} x(x-1)=0 \\ x=0,1 \end{gathered}$$

Therefore, the domain will be $\left\{x\right|x\ne 0,1\}$.

Step 4. Find f ⋅ g and its domain.

We know $f\cdot g=f\left(x\right)\cdot g\left(x\right)$.

Therefore,

$$\begin{gathered} f\cdot g \\ =\left(\frac{x+1}{x-1}\right)\left(\frac{1}{x}\right) \\ =\frac{x+1}{x(x-1)} \end{gathered}$$

The domain will be the set of all real numbers except those that make the denominator zero.

$$\begin{gathered} x(x-1)=0 \\ x=0,1 \end{gathered}$$

Therefore, the domain will be $\left\{x\right|x\ne 0,1\}$.

Step 5. Find f g and its domain.

We know $\frac{f}{g}=\frac{f\left(x\right)}{g\left(x\right)}$.

Therefore,

$$\begin{gathered} \frac{f}{g} \\ =\frac{{\displaystyle \frac{x+1}{x-1}}}{{\displaystyle \frac{1}{x}}} \\ =\frac{x(x+1)}{(x-1)} \end{gathered}$$

The domain will be the set of all real numbers except those that make the denominator zero.

$$\begin{gathered} x-1=0 \\ x=1 \end{gathered}$$

Therefore, the domain will be $\left\{x\right|x\ne 1\}$.