Algebra of functions

Two real functions f and g can be combined point by point, for every x in both their domains:

• (f + g)(x) = f(x) + g(x) and (f – g)(x) = f(x) – g(x).
• (αf)(x) = α f(x) for any real number α.
• (fg)(x) = f(x) g(x).
• (f/g)(x) = f(x)/g(x), but only where g(x) ≠ 0.

Example: f(x) = 3x – 1 and g(x) = x + 2, both on R.

• (f + g)(x) = 4x + 1 and (f – g)(x) = 2x – 3. When subtracting, put g in brackets: 3x – 1 – (x + 2).
• (5f)(x) = 15x – 5.
• (fg)(x) = (3x – 1)(x + 2) = 3x² + 5x – 2.
• (f/g)(x) = (3x – 1)/(x + 2), with x ≠ –2.

Notice that dividing changes the domain. f and g are both defined for every real number, but f/g is not defined at x = –2. Adding, subtracting and multiplying keep the common domain.

Do not confuse (fg)(x), the product f(x) × g(x), with f(g(x)), putting g(x) into f. Here f(g(x)) = 3(x + 2) – 1 = 3x + 5, which is not 3x² + 5x – 2.

Evaluating means replacing x by the given input everywhere in the rule.

• For f(x) = x² – 3x: f(–2) = 4 + 6 = 10, and f(x + 1) = (x + 1)² – 3(x + 1) = x² – x – 2.
• Even a whole expression can go in: f(f(x)) means put f(x) wherever x appears.

Piecewise rules are evaluated piece by piece. Take f(x) = 2x + 1 for x < 0, f(0) = 3, and f(x) = x² for x > 0. Then f(–2) = –3, f(0) = 3 and f(4) = 16: first decide which piece the input belongs to.

Formulas like this describe real situations. For example, a taxi charging ₹50 plus ₹15 per km has fare f(d) = 50 + 15d, and a 12 km ride costs f(12) = ₹230.