Section 1
Where Did Function Arithmetic Come From?
Functions are one of the most important ideas in all of mathematics, but they didn't appear overnight. For centuries, mathematicians worked to develop a clear language for describing how quantities depend on each other. The ability to combine functions using arithmetic operations — adding them together, subtracting one from another, multiplying them, or dividing them — grew naturally from the need to model increasingly complex real-world situations. Think of it this way: once mathematicians had simple functions as "building blocks," they wanted to snap those blocks together to create more intricate models, just like combining ingredients in a recipe to make something new.
~1637 • René Descartes
1694 • Gottfried Wilhelm Leibniz
1748 • Leonhard Euler
f(x) that we still use today. Euler showed how different types of functions — polynomials, trigonometric functions, and exponentials — could be combined through arithmetic to model physical phenomena like vibrating strings and planetary motion.1800s • Cauchy & Weierstrass
Today • Modern Applications
The key question this lesson addresses is straightforward but powerful: if you already know two simple functions, how can you combine them into a single new function that captures a more complex relationship? By the end of this lesson, you'll be able to answer that question with confidence.
Section 2
Core Principles & Definitions
Before you start combining functions, let's make sure the foundation is solid. A function is a rule that takes an input value (usually called x) and produces exactly one output value. You've already worked with functions like f(x) = 2x + 3 (a linear function) and g(x) = x² (a quadratic function). When we talk about combining functions using arithmetic operations, we mean using addition, subtraction, multiplication, or division to create a brand-new function from two existing ones.
Addition of Functions
f and g, you add their outputs for the same input: (f + g)(x) = f(x) + g(x). You simply combine like terms from both expressions.Subtraction of Functions
g away from f: (f − g)(x) = f(x) − g(x). Be careful to distribute the negative sign to every term in g(x).Multiplication of Functions
(f · g)(x) = f(x) · g(x). This often requires the distributive property (like FOIL for binomials).Division of Functions
(f / g)(x) = f(x) / g(x), but only when g(x) ≠ 0. Division by zero is undefined, so some inputs may need to be excluded.Each of these operations takes two functions you already know and produces a new function. The domain (the set of valid inputs) of the new function is usually the overlap of the domains of the original two functions, with the extra restriction for division that the denominator can't be zero.
Key Takeaway
f(x) is a recipe for making lemonade and g(x) is a recipe for making iced tea, then (f + g)(x) is like combining both recipes to make an Arnold Palmer. You're using the same ingredients (the same input x), but the result is something new. Each arithmetic operation gives you a different way to "mix" your function recipes.Section 3
Seeing Function Arithmetic on a Graph
One of the best ways to understand how combining functions works is to see it visually. When you add two functions, the graph of the new function is formed by adding the y-values (the outputs) of the two original functions at every x-value. The diagram below shows this in action. We have a linear function f(x) = x + 1 (shown in cyan) and a quadratic function g(x) = x² (shown in violet). The sum (f + g)(x) = x² + x + 1 is shown in amber. Notice how, at each x-value, the amber curve's height equals the combined heights of the cyan line and the violet parabola.
Look at the dashed lines at x = 1. The cyan function gives f(1) = 2 and the violet function gives g(1) = 1. When you add them, you get (f + g)(1) = 3, which is exactly where the amber curve sits. This "stacking" of outputs happens at every single point along the x-axis. If you understand this visual idea, you understand the core of function addition — and the same logic (with subtraction, multiplication, or division instead of addition) applies to the other operations.
Section 4
The Mathematical Framework
Now let's write down the formal definitions. Suppose you have two functions, f(x) and g(x). Here are the four arithmetic operations you can perform on them.
Sum of Two Functions
Add the outputs — combine like terms in the expressions.
Difference of Two Functions
Subtract the outputs — distribute the negative sign through all terms of g(x).
Product of Two Functions
Multiply the outputs — use the distributive property to expand.
Quotient of Two Functions
Divide the outputs — the domain excludes any x where g(x) = 0.
These definitions might look simple, and in many ways they are. The key skill is careful algebra: making sure you distribute negatives correctly, multiply using FOIL or the distributive property, and identify domain restrictions when dividing. Let's also be precise about what "domain" means in this context. The domain of a combined function is the set of all x-values where both original functions are defined. For the quotient, you must also exclude any x-value that makes the denominator equal to zero.
Here's a quick way to remember the process: you are always performing the arithmetic on the output side — the right side of the equation — while the input stays the same. The input x goes into both functions, and then you combine whatever comes out.
Section 5
Operation-by-Operation Breakdown
Let's look at each operation in more detail with specific function types. Throughout this section, we'll use f(x) = 3x + 2 (a linear function) and g(x) = x² − 1 (a quadratic function) as our running examples.
Notice something important: the sum and difference both produce quadratic functions (the highest power is x²), the product creates a cubic function (degree 3, since degree 1 × degree 2 = degree 3), and the quotient produces a rational function (a fraction with polynomials on top and bottom). This is a general pattern: addition and subtraction preserve the highest degree, multiplication adds the degrees together, and division creates a ratio.
| Operation | Result with f(x) = 3x + 2, g(x) = x² − 1 | Result Type | Domain Restriction |
|---|---|---|---|
| f + g | x² + 3x + 1 | Quadratic | All real numbers |
| f − g | −x² + 3x + 3 | Quadratic | All real numbers |
| f · g | 3x³ + 2x² − 3x − 2 | Cubic | All real numbers |
| f / g | (3x + 2) / (x² − 1) | Rational | x ≠ 1 and x ≠ −1 |
For the division case, g(x) = x² − 1 equals zero when x = 1 or x = −1 (since x² − 1 = (x − 1)(x + 1)). That's why we must exclude those values from the domain. Always check whether the denominator can equal zero and note those restrictions.
Section 6
Worked Example
Let's work through a complete problem step by step. Suppose f(x) = x² + 4x and g(x) = 2x − 5. Find (f − g)(x), (f · g)(3), and state the domain of (f / g)(x).
Complete Worked Example
- 1
Step 1 — Find (f − g)(x)
Write the subtraction definition and substitute the expressions:(f − g)(x) = f(x) − g(x) = (x² + 4x) − (2x − 5)Distribute the negative sign to every term in g(x):= x² + 4x − 2x + 5Combine like terms:4x − 2x = 2x(f − g)(x) = x² + 2x + 5 - 2
Step 2 — Find (f · g)(3)
You could multiply the entire expressions first, or you can take a shortcut: evaluate each function atx = 3separately, then multiply the results.f(3) = (3)² + 4(3) = 9 + 12 = 21g(3) = 2(3) − 5 = 6 − 5 = 1Now multiply:(f · g)(3) = f(3) · g(3) = 21 × 1(f · g)(3) = 21 - 3
Step 3 — State the domain of (f / g)(x)
For the quotient(f / g)(x) = (x² + 4x) / (2x − 5), we need the denominator to not equal zero. Set2x − 5 = 0and solve:2x = 5, sox = 5/2.Domain: All real numbers except x = 5/2 (or x = 2.5) - 4
Step 4 — Interpret the Results
The subtraction produced a quadratic function. The product atx = 3gave us a specific numerical answer, which is a handy shortcut when you only need one value. The quotient can accept any input exceptx = 2.5, where the denominator would be zero and the function is undefined. Notice that we checked each operation separately — the domain restriction only matters for the division.
Section 7
Strengths, Limitations & Common Mistakes
Combining functions through arithmetic is a powerful technique, but there are common mistakes students make. Understanding where things can go wrong will help you avoid errors on homework and tests. Let's compare the operations side by side and highlight what to watch for.
| Feature | Addition / Subtraction | Multiplication | Division |
|---|---|---|---|
| Difficulty | Easiest — just combine like terms | Moderate — requires distribution | Moderate — watch for domain |
| Common mistake | Forgetting to distribute the negative sign (subtraction) | Missing cross terms in FOIL | Forgetting to exclude values where denominator = 0 |
| Result type | Same as highest-degree input | Degree = sum of degrees | Rational function (fraction) |
| Domain change? | Usually no | Usually no | Yes — must exclude zeros of denominator |
| Can evaluate at a point? | Yes: f(a) + g(a) | Yes: f(a) × g(a) | Yes, if g(a) ≠ 0 |
The single most frequent error in function subtraction is forgetting that the negative sign applies to every term in the second function. For example, (x² + 3) − (2x − 7) becomes x² + 3 − 2x + 7, not x² + 3 − 2x − 7. The −7 flips to +7. Always use parentheses when you write out the subtraction, then deliberately distribute the negative to each term.
Key Takeaway
g(x) = 0), everything collapses. Always check whether the "floor" can be zero and note those forbidden values.Section 8
Connection to More Advanced Topics
The arithmetic operations on functions that you're learning now are the starting point for several more advanced ideas you'll encounter in later math courses. Understanding these connections can help you see why this skill matters beyond just this chapter.
| This Lesson | Where It Leads | Course |
|---|---|---|
| Adding/subtracting functions | Piecewise functions and modeling with combined formulas | Algebra 2 |
| Multiplying functions | Polynomial factoring, area under curves | Algebra 2 / Precalculus |
| Dividing functions | Rational functions, asymptotes, limits | Precalculus / Calculus |
| Evaluating (f ± g)(a) at a point | Composition of functions: f(g(x)) | Algebra 2 |
| Domain restrictions from division | Continuity, removable discontinuities | Calculus |
One idea that's especially worth previewing is function composition, written f(g(x)). While arithmetic operations combine the outputs of two functions side by side, composition chains them — the output of one function becomes the input of the next. That's a related but distinct idea you'll study soon. For now, just remember: arithmetic combines outputs, and composition chains inputs and outputs. The fact that you can build complex functions from simple ones, whether by arithmetic or by composition, is what makes functions such a flexible and powerful tool in mathematics.
Section 9
Practice Problems
Test your understanding with these five problems. They start with a conceptual check and build up to a challenge problem. Try each one before clicking to reveal the answer.
Question 1: Problem 1 — Conceptual
(f + g)(x). Why does the result count as a new function?Show answer
(f + g)(x) means you take the same input x, plug it into both f and g separately to get their individual outputs, and then add those outputs together. The result is a new function because, for every input value x, there is exactly one output — the sum of f(x) and g(x). Since it follows the rule of assigning each input to exactly one output, it qualifies as a function on its own.Question 2: Problem 2 — Basic Calculation
f(x) = 5x − 3 and g(x) = x + 7. Find (f + g)(x) and evaluate (f + g)(2).Show answer
(f + g)(x) = f(x) + g(x) = (5x − 3) + (x + 7) = 5x − 3 + x + 7 = 6x + 4
Now evaluate at x = 2: (f + g)(2) = 6(2) + 4 = 12 + 4 = 16.
You could also check this by computing f(2) = 5(2) − 3 = 7 and g(2) = 2 + 7 = 9, and then 7 + 9 = 16. Both methods give the same answer.Question 3: Problem 3 — Intermediate
f(x) = x² + 2x and g(x) = 3x − 4. Find (f − g)(x) and (f · g)(x).Show answer
(f − g)(x) = (x² + 2x) − (3x − 4) = x² + 2x − 3x + 4 = x² − x + 4. Remember to distribute the negative: −(3x − 4) = −3x + 4.
Multiplication: (f · g)(x) = (x² + 2x)(3x − 4). Use the distributive property:
= x²(3x) + x²(−4) + 2x(3x) + 2x(−4) = 3x³ − 4x² + 6x² − 8x = 3x³ + 2x² − 8x
Combining like terms: −4x² + 6x² = 2x². The final product is a cubic function with degree 3.Question 4: Problem 4 — Applied / Multi-Step
R(x) = 15x (dollars earned from selling x items) and its cost with C(x) = 3x + 40 (dollars spent to produce x items plus a $40 fixed cost). The profit function is P(x) = R(x) − C(x). Find P(x), then determine how many items must be sold to earn at least $80 in profit.Show answer
P(x) = R(x) − C(x) = 15x − (3x + 40) = 15x − 3x − 40 = 12x − 40
Step 2: Set the profit greater than or equal to $80 and solve:
12x − 40 ≥ 80 → 12x ≥ 120 → x ≥ 10
The business must sell at least 10 items to earn $80 or more in profit. This problem demonstrates a real-world use of function subtraction: profit is always revenue minus cost.Question 5: Problem 5 — Critical Thinking / Synthesis
f(x) = x + 3 and g(x) = x² − 9. Write the quotient (g / f)(x), simplify if possible, and state the domain. Then explain: does the simplified version have the same domain as the unsimplified version? Why or why not?Show answer
(g / f)(x) = (x² − 9) / (x + 3)
Factor the numerator: x² − 9 = (x − 3)(x + 3) (difference of squares).
So (g / f)(x) = (x − 3)(x + 3) / (x + 3).
Simplify: Cancel the common factor (x + 3) to get x − 3.
Domain: The original quotient requires x + 3 ≠ 0, which means x ≠ −3. Even though the simplified form x − 3 looks like it works for all values of x, the domain is still all real numbers except x = −3. That's because the original function was undefined at x = −3, and simplifying doesn't change the domain. The simplified form just has a "hole" at x = −3 where the function doesn't exist. This is an important concept you'll explore more deeply in later courses under the name "removable discontinuity."Summary
Lesson Summary
Summary
In this lesson, you learned how to combine standard function types using arithmetic operations. Given two functions f(x) and g(x), you can create new functions by adding their outputs (f + g)(x) = f(x) + g(x), subtracting them (f − g)(x) = f(x) − g(x), multiplying them (f · g)(x) = f(x) · g(x), or dividing them (f / g)(x) = f(x) / g(x) when g(x) ≠ 0. Each operation follows specific algebraic rules: addition and subtraction require combining like terms (with careful attention to distributing negatives), multiplication requires the distributive property, and division requires checking for domain restrictions wherever the denominator equals zero.
These operations produce new function types: adding or subtracting keeps the highest degree, multiplying creates a function whose degree is the sum of the original degrees, and dividing produces a rational function. You saw how this works visually on a graph — where the new function's height at each point is determined by combining the heights of the original functions — and through a real-world business application involving profit = revenue − cost. These skills form the foundation for more advanced topics like function composition, rational expressions, and calculus.