6.3 Function Operations, Composition, and Inverses

6.3 Function Operations, Composition, and Inverses

Once f and g exist, AAF asks you to combine them and then to invert them. Table 11’s “creating functions with function notation” and “evaluating” both cover (f+g)(x) and f(g(x)), and the top Skills Insight band, 276–300, evaluates logarithmic equations — the named inverse of exponentials.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Arithmetic operations on functions

For functions f and g and a shared input x in both domains:

The product notation (fg)(x) is not composition. Composition uses f(g(x)) or (f ∘ g)(x).

Worked numbers

Let f(x) = x^2 and g(x) = x − 3.

• (f+g)(x) = x^2 + x − 3
• (f−g)(x) = x^2 − (x − 3) = x^2 − x + 3 — the minus sign distributes across both terms of g
• (fg)(x) = x^2(x − 3) = x^3 − 3x^2
• (f/g)(x) = x^2 / (x − 3), domain all reals except x = 3

Evaluate at 5: (f+g)(5) = 25 + 2 = 27, which is the same as f(5)+g(5) = 25 + 2. (f/g)(5) = 25/2. (f/g)(3) is undefined, even though f(3) = 9 exists. Domain restrictions on the quotient are not optional decorations; they are part of the answer.

A second trap: (f−g)(x) is not f(g(x)) and not f(x) − g. If you subtract the formulas, parenthesize g. If g(x) = √(x − 1), then (f+g) is defined only for x ≥ 1, the stricter of the two domains, even though f(x) = x^2 was happy with every real x.

AAF will also hand you two short formulas and ask for a single number such as (f+g)(−2). Compute each output, then add: if f(x) = 3x and g(x) = x^2 − 1, then f(−2) = −6, g(−2) = 3, and (f+g)(−2) = −3.

Composition: f(g(x)) versus g(f(x))

Composition f(g(x)) means: take x, apply g, then feed that output into f. The inner function runs first. g(f(x)) reverses the order. They are equal only in special cases (for example when they are inverses of each other).

Worked example

Let f(x) = x^2 + 1 and g(x) = 3x − 2.

f(g(x)) = f(3x − 2) = (3x − 2)^2 + 1 = 9x^2 − 12x + 4 + 1 = 9x^2 − 12x + 5.

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

These are different polynomials. Check at x = 1: g(1) = 1, f(1) = 2. Then f(g(1)) = f(1) = 2 and 9 − 12 + 5 = 2. Meanwhile g(f(1)) = g(2) = 4 and 3 + 1 = 4. Same input, different composed outputs.

Numeric composition without simplifying is common. If a table gives g(2) = 4 and f(4) = 17, then f(g(2)) = 17 even when you do not have formulas. Read inside to outside: find g(2) first, then look that number up in f. Reversing the table lookup is how g(f(2)) is born by accident.

Domain of f(g(x)): x must be in the domain of g, and g(x) must be in the domain of f. If f(x) = √x and g(x) = x − 5, then f(g(x)) = √(x − 5) needs x ≥ 5. If g is 1/x, then x ≠ 0 in addition to whatever f requires of g(x).

Inverses: swap, solve, check

The inverse f^{-1} undoes f. By definition, f(f^{-1}(x)) = x and f^{-1}(f(x)) = x on the appropriate sets. The superscript −1 is not an exponent: f^{-1}(x) is not 1/f(x). Reciprocal is 1/f(x) or [f(x)]^{-1} written with the whole output in the denominator.

Algebraic method

1. Replace f(x) with y.
2. Swap x and y.
3. Solve for y.
4. Rename y as f^{-1}(x).
5. Check f(f^{-1}(x)) = x.

Worked: f(x) = 2x − 5.

• y = 2x − 5
• x = 2y − 5
• x + 5 = 2y
• y = (x + 5)/2

so f^{-1}(x) = (x + 5)/2.

Check: f(f^{-1}(x)) = 2 · ((x+5)/2) − 5 = x + 5 − 5 = x. And f^{-1}(f(x)) = (2x − 5 + 5)/2 = x.

Evaluate without the formula if the item only needs a number. f^{-1}(9) means “what input to f produces 9?” Solve 2x − 5 = 9, x = 7. From a table, look up the output 9 and read the matching input. From a graph, find the height y = 9 and read x. That is exactly the inverse evaluation AAF expects.

Another numeric check: f(4) = 3, so f^{-1}(3) = 4. If you are given f^{-1}(6) = −1, then f(−1) = 6. Inverse pairs swap coordinates.

One-to-one and the horizontal line test

An inverse function exists only when f is one-to-one: different inputs give different outputs. The horizontal line test is the graph version. If any horizontal line meets the graph twice, two inputs share an output, and you cannot uniquely reverse it.

• f(x) = 2x − 5 passes (a non-horizontal line). Inverse exists.
• f(x) = x^3 passes. Inverse is the cube root.
• f(x) = x^2 on all reals fails: f(3) = f(−3) = 9. No inverse function on all reals. Restrict the domain to x ≥ 0; then f^{-1}(x) = √x.
• f(x) = |x| fails on all reals for the same two-preimage reason.

AAF will not usually ask you to write a restricted-domain inverse from scratch, but it will expect you not to “invert” x^2 as ±√x and call that a function. Two outputs for one input violate the definition from Graphs of Functions, Vertical Line Test, Maxima and Minima.

Graphs of inverses are reflections of each other across the line y = x. If (3, 11) is on f, then (11, 3) is on f^{-1}. Vertex and intercept graphs of quadratics, which usually fail the horizontal line test unless the domain is cut, are in Quadratic Forms, Graphs, Vertex, and Intercepts.