Adding Functions

Adding functions is exactly what it sounds like: you take two functions and combine their output values by addition. If you already know how to add polynomials or rational expressions, adding functions uses the same skills, just written with function notation.

Given two functions \(f(x)\) and \(g(x)\), the sum function is written \((f+g)(x)\), and the rule is:

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

In other words, to find the sum function, you substitute the expressions for \(f(x)\) and \(g(x)\) into the right-hand side, then simplify by combining like terms.

The most common case you will meet first is adding two polynomial functions. The process has three steps:

1. Write out \(f(x) + g(x)\) with both expressions in full.
2. Remove any parentheses (distribute a negative sign only if you were subtracting, not adding).
3. Combine like terms so the answer is in simplest form.

Example 1: Let \(f(x) = 3x^2 - 2x + 1\) and \(g(x) = x^2 + 5x - 4\). Find \((f+g)(x)\).

\((f+g)(x) = (3x^2 - 2x + 1) + (x^2 + 5x - 4)\)

\(= 3x^2 + x^2 - 2x + 5x + 1 - 4\)

\(= 4x^2 + 3x - 3\)

You can also evaluate a sum function at a specific input by either adding first and then substituting, or substituting into each function separately and adding the results, both give the same answer.

Example 2: Using the same \(f\) and \(g\) above, find \((f+g)(2)\).

Using the simplified sum \(4x^2 + 3x - 3\):

\((f+g)(2) = 4(2)^2 + 3(2) - 3 = 16 + 6 - 3 = 19\)

Check it the other way: \(f(2) = 3(4) - 4 + 1 = 9\) and \(g(2) = 4 + 10 - 4 = 10\), so \(f(2) + g(2) = 9 + 10 = 19\). Both methods match.

Adding rational functions works the same way as other function operations, but you also need to handle fractions, which means finding a common denominator, just as you would for addition of rational expressions.

Example 3: Let \(f(x) = \frac{1}{x}\) and \(g(x) = \frac{2}{x+3}\). Find \((f+g)(x)\).

The denominators are \(x\) and \(x+3\), so the common denominator is \(x(x+3)\):

\((f+g)(x) = \frac{1}{x} + \frac{2}{x+3} = \frac{x+3}{x(x+3)} + \frac{2x}{x(x+3)}\)

\(= \frac{x + 3 + 2x}{x(x+3)} = \frac{3x+3}{x(x+3)}\)

When you add functions, the domain of the result is the intersection of the domains of the two original functions, meaning every input value that works in both \(f(x)\) and \(g(x)\). If either original function has a restriction, that restriction carries over into the sum. To review how to identify these restrictions in the first place, see how to write domain and range.

In Example 3, \(f(x) = \frac{1}{x}\) is undefined at \(x = 0\), and \(g(x) = \frac{2}{x+3}\) is undefined at \(x = -3\). So the sum function \((f+g)(x) = \frac{3x+3}{x(x+3)}\) is defined for all real numbers except \(x = 0\) and \(x = -3\), even though the simplified expression does not obviously show those restrictions at a glance.

To add functions: write \((f+g)(x) = f(x) + g(x)\), substitute both expressions, and simplify by combining like terms or finding a common denominator when fractions are involved. Always carry over any domain restrictions from the original functions into your final answer. This same substitute-and-simplify approach also underlies other function operations you will meet later, such as subtraction, multiplication, and the difference quotient.

Related lessons

• Algebra Help