Chapter 2, Problem 12RE

In Problems 12-14, find $f\text{ + g, }f-g,f\cdot g,\text{ and }\frac{f}{g}$ for each pair of functions. State the domain of each of these functions.

$f\left(x\right)\text{ = 2 }-x;\text{ g}\left(x\right)\text{ = 3}x\text{ + 1}$

To determine

To find: The pair of function from the provided functions.

Answer to Problem 12RE

Solution:

a. The domain of the pair $f+g$ is the set of all real numbers.

b. The domain of the pair $f-g$ is the set of all real numbers.

c. The domain of the pair $f\circ g$ is the set of all real numbers.

d. The domain of the pair $\frac{f}{G}$ is $\left\{x\right|x\ne \frac{-1}{3},x\in ℝ\}$.

Explanation of Solution

Given:

It is provided in the problem that the function $f\left(x\right)=2-x$ and $g\left(x\right)=3x+1$.

Formula used:

A quadratic equation is any equation/function with a degree of 2 that can be written in the form $y=a{x}^2+bx+c$, where $a,b$ and $c$ are real numbers, and a does not equal to 0. Its graph is called a parabola. The constants $a,b$, and $c$ are called the parameters of the equation. The values of $a,b$ and $c$ determine the shape and position of the parabola.

The domain of a function is the set of all real values of $x$ that will give real values for $y$. The range of a function is the set of all real values of $y$ that you can get by plugging real numbers into $x$.

Calculation:

a. For $f+g$, $f\left(x\right)+g\left(x\right)=2-x+3x+1=2x+1$. Here the function $f+g$ is defined 2 multiplied by $x$ and added 1 with it. The input of the function $f+g$ takes all the real values in the real line. Therefore, the domain of the pair $f+g$ is the set of all real numbers.

b. For $f-g$, $f\left(x\right)-g\left(x\right)=2-x-3x-1=1-4x$. Here the function $f-g$ is defined $-4$ multiplied by $x$ and added 1 with it. The input of the function $f-g$ takes all the real values in the real line. Therefore, the domain of the pair $f-g$ is the set of all real numbers.

c. For $f\circ g$, $f\left(x\right)\cdot g\left(x\right)=\left(2-x\right)\times \left(3x+1\right)=6x+2-3x^2-x=-3x^2+5x+2$. Here the function $f\circ g$ is a quadratic equation. By the definition of domain of the quadratic equation, the domain takes all the real values. Therefore, the domain of the pair $f\circ g$ is the set of all real numbers.

d. For $\frac{f}{G}$, $\frac{f\left(x\right)}{g\left(x\right)}=\frac{2-x}{3x+1}$. The function $\frac{f}{G}$ says to divide $2-x$ by $3x+1$. Since division by 0 is not defined, the denominator $3x+1$ can never be 0, so $x$ can never equal $\frac{-1}{3}$. The domain of the pair $\frac{f}{G}$ is $\left\{x\right|x\ne \frac{-1}{3},x\in ℝ\}$.