DWG NO. 1.5 — Lesson 5 of 6

Combining Functions

Unit 1: Functions and Their Graphs · ~25 min

Objective Perform sum, difference, product, quotient, and composition operations on functions, and find the domain of each result.

Arithmetic on functions

Two functions can be combined the same way two numbers can — added, subtracted, multiplied, or divided. Each combination is itself a new function:

(f + g)(x) = f(x) + g(x)
(f − g)(x) = f(x) − g(x)
(f · g)(x) = f(x) · g(x)
(f / g)(x) = f(x) / g(x), provided g(x) ≠ 0

The domain of a sum, difference, or product is whatever both f and g's domains have in common. The quotient has that same restriction, plus it must also exclude any x where g(x) = 0.

Composition

Composition is different from the four operations above — it's not arithmetic on outputs, it's chaining two functions so one's output becomes the other's input. (f \(\circ\) g)(x), read "f of g of x," means: run x through g first, then run that result through f.

(f \(\circ\) g)(x) = f(g(x)). Order matters — (f \(\circ\) g)(x) and (g \(\circ\) f)(x) are generally different functions.

Composition (f \(\circ\) g)(x) — x goes through g first, and g's output becomes f's input

Worked Example 1 · Sum and product

ProblemLet f(x) = x² + 1 and g(x) = 2x − 3. Find (f+g)(x) and (f·g)(x).

1(f+g)(x) = (x²+1) + (2x−3) = x² + 2x − 2.

2(f·g)(x) = (x²+1)(2x−3) = 2x³ − 3x² + 2x − 3.

(f+g)(x) = x² + 2x − 2 · (f·g)(x) = 2x³ − 3x² + 2x − 3

Worked Example 2 · Composition both ways

ProblemLet f(x) = 2x + 1 and g(x) = x². Find (f \(\circ\) g)(x) and (g \(\circ\) f)(x).

1(f \(\circ\) g)(x) = f(g(x)) = f(x²) = 2(x²) + 1 = 2x² + 1.

2(g \(\circ\) f)(x) = g(f(x)) = g(2x+1) = (2x+1)² = 4x² + 4x + 1.

(f \(\circ\) g)(x) = 2x² + 1 · (g \(\circ\) f)(x) = 4x² + 4x + 1 — not equal, so order matters

Worked Example 3 · Domain of a quotient

ProblemLet f(x) = x + 3 and g(x) = x − 5, both with domain all real numbers. Find the domain of (f/g)(x).

1(f/g)(x) = (x+3)/(x−5) needs g(x) ≠ 0: x − 5 ≠ 0, so x ≠ 5.

Domain of (f/g)(x): (−∞, 5) ∪ (5, ∞)

Guided practice

1.Let f(x) = 3x − 1 and g(x) = x + 4. Find (f − g)(x).
Show answer

(f−g)(x) = (3x−1) − (x+4) = 2x − 5. (f−g)(x) = 2x − 5
2.Let f(x) = x² and g(x) = x − 1. Find (f \(\circ\) g)(x).
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(f \(\circ\) g)(x) = f(g(x)) = f(x−1) = (x−1)² = x² − 2x + 1. (f \(\circ\) g)(x) = x² − 2x + 1
3.Using the same f and g from problem 2, find (g \(\circ\) f)(x), and confirm it's different from (f \(\circ\) g)(x).
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(g \(\circ\) f)(x) = g(f(x)) = g(x²) = x² − 1. (g \(\circ\) f)(x) = x² − 1, which is different from x² − 2x + 1 — confirming order matters.
4.Let f(x) = \(\sqrt{x}\) (domain x ≥ 0) and g(x) = x − 4 (domain all reals). Find the domain of (f+g)(x).
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The sum's domain is where both domains overlap: f needs x ≥ 0, g allows everything. Domain: [0, ∞)
5.Let f(x) = 1/x and g(x) = x + 2. Find (f/g)(x) and its domain.
Show answer

(f/g)(x) = (1/x) ÷ (x+2) = \(\frac{1}{[x}\)(x+2)]. Exclude x = 0 (f's own restriction) and x = −2 (where g(x) = 0). Domain: (−∞, −2) ∪ (−2, 0) ∪ (0, ∞)

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