Chapter 1.5, Problem 54E

(a)

To determine

To calculate: To determine the domain of the function $f$

Answer to Problem 54E

Domain is $R-\left\{-1,1\right\}$

Explanation of Solution

Given: $f\left(x\right)=\frac{3}{x^2-1}$

Formula Used:

Domain of a function is the set of all possible inputs for the function

Calculation:

Function is given as follows:

  $f\left(x\right)=\frac{3}{x^2-1}$

For $f\left(x\right)$ to be defined

  $\begin{array}{l}{x}^2-1\ne 0\\ x^2\ne 1\\ x\ne \pm 1\end{array}$

Domain of $f\left(x\right)$ is set of all real numbers except $-1,1$

Thus, domain is $R-\left\{-1,1\right\}$

Conclusion:

Hence, domain is $R-\left\{-1,1\right\}$

(b)

To determine

To calculate: To determine the domain of the function $g$

Answer to Problem 54E

Domain is $R$

Explanation of Solution

Given: $g\left(x\right)=x+1$

Formula Used:

Domain of a function is the set of all possible inputs for the function

Calculation:

Function is given as follows:

  $g\left(x\right)=x+1$

Domain of $g\left(x\right)$ is set of all real numbers

Thus, domain is $R$

Conclusion:

Hence, domain is $R$

(c)

To determine

To calculate: To determine the domain of the $f\circ g$

Answer to Problem 54E

Domain is $R-\left\{0,-2\right\}$

Explanation of Solution

Given: $f\left(x\right)=\frac{3}{x^2-1}$ and $g\left(x\right)=x+1$

Formula Used:

Composition of functions is when one function is inside of another function.

The notation used for the composition of functions looks like this, $\left(f\circ g\right)\left(x\right)$ .

The composition of the function f with g is defined as follows:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ , notice that in the case the function g is inside of the function

Also, Domain of a function is the set of all possible inputs for the function

Calculation:

Given function as follows:

  $f\left(x\right)=\frac{3}{x^2-1}$ and $g\left(x\right)=x+1$

  $f\circ g$ is calculated as follows:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $g\left(x\right)=x+1$

  $\left(f\circ g\right)\left(x\right)=f\left(x+1\right)$

Now substitute $f\left(x\right)=\frac{3}{x^2-1}$

  $\begin{array}{l}\left(f\circ g\right)\left(x\right)=\frac{3}{{\left(x+1\right)}^2-1}\\ \left(f\circ g\right)\left(x\right)=\frac{3}{x^2+2x+1-1}\\ \left(f\circ g\right)\left(x\right)=\frac{3}{x^2+2x}\end{array}$

For $\left(f\circ g\right)\left(x\right)$ to be defined

  $\begin{array}{l}{x}^2+2x\ne 0\\ x\left(x+2\right)\ne 0\\ x\ne 0,-2\end{array}$

Domain of $\left(f\circ g\right)\left(x\right)$ is set of all real numbers except $0,-2$

Thus, domain is $R-\left\{0,-2\right\}$

Conclusion:

Hence, domain is $R-\left\{0,-2\right\}$