Addition and Subtraction of Functions

Combining Functions

Imagine you have two function machines, let's call them machine $f$ and machine $g$. Each machine has its own rules, which are its function ($f(x)$ and $g(x)$) and the raw materials it can process (its domain, $D_f$ and $D_g$). We can combine these two machines to create a new machine using addition or subtraction operations.

Addition of Two Functions

If we want to add function $f$ and function $g$, we simply add the results from each function for the same value of $x$. The result is a new function we call $(f+g)$.

Important note: The combined machine $(f+g)$ can only process raw materials (values of $x$) that can be processed by both original machines, $f$ and $g$. So, the domain (domain of origin) of the function $(f+g)$ is the intersection of the domain of $f$ and the domain of $g$.

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:

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

Step 1: Determine the resulting function from addition

Step 2: Determine the domain of the resulting function

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

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 $(f-g)$.

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.

Example of Subtraction

We use the same functions as in the addition example:

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

Step 1: Determine the resulting function from subtraction

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:

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 $D_g = \{x | x \in \mathbb{R}\}$.

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

Answer Key

1. Finding $(f+g)(x)$:

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

   $$= (2x - 1) + (x^2 + 3)$$

   $$= x^2 + 2x + 2$$

   Finding Domain $D_{f+g}$:

   $$D_{f+g} = D_f \cap D_g$$

   $$= \{x | x \in \mathbb{R}\} \cap \{x | x \in \mathbb{R}\}$$

   $$= \{x | x \in \mathbb{R}\}$$

   So, $(f+g)(x) = x^2 + 2x + 2$ with domain all real numbers.

2. Finding $(f-g)(x)$:

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

   $$= (2x - 1) - (x^2 + 3)$$

   $$= 2x - 1 - x^2 - 3$$

   $$= -x^2 + 2x - 4$$

   Finding Domain $D_{f-g}$:

   $$D_{f-g} = D_f \cap D_g$$

   $$= \{x | x \in \mathbb{R}\} \cap \{x | x \in \mathbb{R}\}$$

   $$= \{x | x \in \mathbb{R}\}$$

   So, $(f-g)(x) = -x^2 + 2x - 4$ with domain all real numbers.

3. Calculating $(f+g)(2)$:

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

   $$(f+g)(2) = (2)^2 + 2(2) + 2$$

   $$= 4 + 4 + 2$$

   $$= 10$$

4. Calculating $(f-g)(-1)$:

   We use the result from number 2: $(f-g)(x) = -x^2 + 2x - 4$

   $$(f-g)(-1) = -(-1)^2 + 2(-1) - 4$$

   $$= -(1) - 2 - 4$$

   $$= -7$$