Algebra of Real-Valued Functions: Operations & Examples

Instructor: Laura Pennington

Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.

This lesson will define real-valued functions using a real-world example. We will then discuss the operations involved in algebra of real-valued functions and look at examples of using each operation on functions.

Real-Valued Functions

Suppose you are the CEO of a successful company, and you are presenting the accounting records for the past year at a meeting. You explain that your company's revenue, or the money coming in, can be modeled using the following formula:

R(x) = -0.01x2 + 3.5x, where x is the number of products produced and sold by the company, and R is in thousands of dollars.

You also display a formula that can model your company's cost, or the amount of money going out.

C(x) = 0.00007x3 - 0.01x2 + 1.26x + 84, where x is the number of products produced and sold by the company, and C is in thousands of dollars.

In mathematics, we call these formulas real-valued functions, or more commonly, just functions.


A function is a rule that relates an input to exactly one output. The term 'is a function of' can be thought of as 'is determined by'. A real-valued function is a function with outputs that are real numbers. Consider our revenue and cost functions. Both cost and revenue are determined by how many products your company produces and sells, so cost and revenue are functions of the number of products produced and sold. They are also both real-valued functions because their outputs are real numbers.

Algebra of Real-Valued Functions

Your employees at the meeting ask if you could find a formula for how much money your company is actually making. In other words, they want to find a formula for your company's profit. Let's see… revenue equals money in, and cost equals money out. Ah-ha! If we subtract the cost from the revenue, we will have our profit. We have a formula for revenue, R(x), and for cost, C(x). Therefore, we can find a formula for profit by subtracting C(x) from R(x).


We get that P(x) = R(x) - C(x). Subtracting the cost function from the revenue function like this is an example of algebra of real-valued functions. Algebra of real-valued functions involves adding, subtracting, multiplying, and dividing real-valued functions, and the rules for each operation are as follows:

  • Addition: (f + g)(x) = f(x) + g(x)
  • Subtraction: (f - g)(x) = f(x) - g(x)
  • Multiplication: (fg)(x) = f(x) ⋅ g(x)
  • Division: (f / g)(x) = f(x) / g(x), where g(x) ≠ 0

If you're thinking that performing these operations on functions looks to be the same as performing them on numbers, you're basically right! It's just a little different because we're working with algebraic expressions rather than just numbers.

For example, consider your company's profit again. To find the profit formula, we just subtract your cost function from your revenue function.


We get that your profit function is P(x) = -0.00007x3 + 2.24x - 84. Huh, this algebra of real-valued functions isn't so hard! Let's look at examples of each of the operations on functions.


Suppose we have the following functions:

  • r(x) = x2 + 4x + 3
  • s(x) = x + 1
  • t(x) = 5x - 17

We want to find r + s, s - t, st, and r / s. Well, nothing to do but get started, so let's do just that! We'll start with r + s. Using our rules, we know that (f + g)(x = f(x) + g(x), so we just need to add the two functions together.


We get that (r + s)(x) = x2 + 5x + 4. On to the next one: s - t. Our subtraction rule for functions states that (f - g)(x) = f(x) - g(x), so let's subtract these functions!


We get that (s - t)(x) = -4x + 18. So far, so good. Next up is st. We use our multiplication rule for functions which states that (fg)(x) = f(x) ⋅ g(x).


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