Addition and Subtraction of Functions

Combining Functions

Two functions can be combined point by point. For the same input $x$ , evaluate $f(x)$ and $g(x)$ , then add or subtract those two outputs. The input must belong to both domains, $D_f$ and $D_g$ .

Function Addition Visualization

Observe how the lines

f(x)=x

and

g(x)=2

are added to become

(f+g)(x)=x+2

Addition of Two Functions

The sum function adds the two outputs at the same input:

(f+g)(x) = f(x) + g(x)

The sum $(f+g)$ is defined only where both original functions are defined. Its domain is therefore the intersection of their domains:

D_{f+g} = D_f \cap D_g

This means that $x$ must be a member of $D_f$ AND also a member of $D_g$ .

Example of Addition

Suppose we have two functions:

$f(x) = x^2$ , with domain $D_f = \{x | x \in \mathbb{R}\}$ (all real numbers).
$g(x) = \sqrt{x+2}$ , with domain $D_g = \{x | x \ge -2, x \in \mathbb{R}\}$ (all real numbers greater than or equal to $-2$ , because the radicand $x+2$ cannot be negative).

Step 1: Determine the resulting function from addition

\begin{aligned} (f+g)(x) &= f(x) + g(x) \\ &= x^2 + \sqrt{x+2} \end{aligned}

Step 2: Determine the domain of the resulting function

We find the intersection of $D_f$ and $D_g$ :

\begin{aligned} &D_{f+g} = D_f \cap D_g \\ &= \{x | x \in \mathbb{R}\} \cap \{x | x \ge -2, x \in \mathbb{R}\} \end{aligned}

So, the resulting function from addition is $(f+g)(x) = x^2 + \sqrt{x+2}$ with domain $\{x | x \ge -2, x \in \mathbb{R}\}$ .

Subtraction of Two Functions

The process is similar to addition. To subtract function $g$ from function $f$ , we subtract the result of $g(x)$ from $f(x)$ for the same value of $x$ . The result is a new function $(f-g)$ .

(f-g)(x) = f(x) - g(x)

Its domain is also the same as for addition, namely the intersection of the domain of $f$ and the domain of $g$ . Why? Because again, the value of $x$ must be processable by both initial functions before it can be subtracted.

D_{f-g} = D_f \cap D_g

Example of Subtraction

We use the same functions as in the addition example:

$f(x) = x^2$ , $D_f = \{x | x \in \mathbb{R}\}$
$g(x) = \sqrt{x+2}$ , $D_g = \{x | x \ge -2, x \in \mathbb{R}\}$

Step 1: Determine the resulting function from subtraction

\begin{aligned} (f-g)(x) &= f(x) - g(x) \\ &= x^2 - \sqrt{x+2} \end{aligned}

Step 2: Determine the domain of the resulting function Its domain is the same as the domain of the addition result because the intersection rule is the same:

\begin{aligned} &D_{f-g} = D_f \cap D_g \\ &= \{x | x \ge -2, x \in \mathbb{R}\} \end{aligned}

So, the resulting function from subtraction is $(f-g)(x) = x^2 - \sqrt{x+2}$ with domain $\{x | x \ge -2, x \in \mathbb{R}\}$ .

Practice Problems

Given the functions $f(x) = 2x - 1$ with $D_f = \{x | x \in \mathbb{R}\}$ and function $g(x) = x^2 + 3$ with $D_g = \{x | x \in \mathbb{R}\}$ .

Determine $(f+g)(x)$ and its domain $D_{f+g}$ .
Determine $(f-g)(x)$ and its domain $D_{f-g}$ .
Calculate the value of $(f+g)(2)$ .
Calculate the value of $(f-g)(-1)$ .

Answer Key

Finding $(f+g)(x)$ :
$(f+g)(x) = f(x) + g(x)$
Finding Domain $D_{f+g}$ :
$D_{f+g} = D_f \cap D_g$
So, $(f+g)(x) = x^2 + 2x + 2$ with domain all real numbers.
Finding $(f-g)(x)$ :
$(f-g)(x) = f(x) - g(x)$
Finding Domain $D_{f-g}$ :
$D_{f-g} = D_f \cap D_g$
So, $(f-g)(x) = -x^2 + 2x - 4$ with domain all real numbers.
Calculating $(f+g)(2)$ :

We use the result from number 1: $(f+g)(x) = x^2 + 2x + 2$
$(f+g)(2) = (2)^2 + 2(2) + 2$
Calculating $(f-g)(-1)$ :

We use the result from number 2: $(f-g)(x) = -x^2 + 2x - 4$
$\begin{aligned} &(f-g)(-1) \\ &= -(-1)^2 + 2(-1) - 4 \end{aligned}$