In short

Given two functions f and g, you can build new functions by adding them (f+g), subtracting (f-g), multiplying (f \cdot g), or dividing \left(\frac{f}{g}\right). In each case, the new function's domain is the intersection of the original domains — with one extra exclusion for division: you must also throw out every x where g(x) = 0.

A phone plan charges you ₹199 per month as a fixed fee, plus ₹1.5 per GB of data you use. Your monthly bill has two parts: a constant function f(x) = 199 and a linear function g(x) = 1.5x, where x is the number of gigabytes. Your total bill is the sum of these two: (f + g)(x) = 199 + 1.5x.

That is function addition — take two functions and add their outputs at each input. The idea is natural. You do it every time you combine two cost components, merge two signals, or layer two effects. The point of this article is to make the idea precise, handle the domain carefully, and catalogue the properties.

The four operations

Let f and g be two real-valued functions. At each input x, you can combine their outputs using the four arithmetic operations.

Algebra of Functions

Given functions f: D_f \to \mathbb{R} and g: D_g \to \mathbb{R}:

Sum: (f + g)(x) = f(x) + g(x)

Difference: (f - g)(x) = f(x) - g(x)

Product: (f \cdot g)(x) = f(x) \cdot g(x)

Quotient: \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}

The domain of the sum, difference, and product is D_f \cap D_g — the set of inputs where both functions are defined.

The domain of the quotient is D_f \cap D_g \setminus \{x : g(x) = 0\} — additionally exclude every x that makes the denominator zero.

The first three are straightforward: if both machines accept the input, you can combine their outputs. The quotient has the extra wrinkle that even if both functions are defined, division by g(x) = 0 is forbidden.

There is also scalar multiplication: for any constant c \in \mathbb{R}, (cf)(x) = c \cdot f(x), with the same domain as f.

Diagram showing two function machines combining outputsAn input x enters two function boxes labelled f and g. The outputs f(x) and g(x) feed into an operation box showing plus, minus, times, or divide. The final output is the combined value. f g x f(x) g(x) + − × ÷ (f ⊕ g)(x)
Two function machines, $f$ and $g$, both receive the same input $x$. Their outputs are combined by an arithmetic operation to produce the output of the new function.

Domain of combined functions

The domain rule is the most important part. If f(x) = \sqrt{x} (domain [0, \infty)) and g(x) = \sqrt{4-x} (domain (-\infty, 4]), then:

  • Domain of f + g: [0, \infty) \cap (-\infty, 4] = [0, 4].
  • Domain of f \cdot g: same — [0, 4].
  • Domain of \frac{f}{g}: start with [0, 4], then exclude x where g(x) = 0. Since \sqrt{4-x} = 0 when x = 4, exclude 4. Domain: [0, 4).
Number lines showing domain intersection of two functionsThree number lines stacked vertically. First: domain of f is 0 to infinity with a solid dot at 0. Second: domain of g is negative infinity to 4 with a solid dot at 4. Third: their intersection is 0 to 4, shaded, with solid dots at both ends. A hollow dot at 4 is shown separately for the quotient domain. D_f 0 4 D_g 0 4 D_f ∩ D_g 0 4 domain of f + g, f − g, f · g
The domain of a combined function is the intersection of the individual domains. $D_f = [0, \infty)$ and $D_g = (-\infty, 4]$ overlap on $[0, 4]$. For the quotient $f/g$, you additionally exclude $x = 4$ where $g(x) = 0$.

Here is a second example to lock in the pattern. Let f(x) = \frac{1}{x} (domain \mathbb{R} \setminus \{0\}) and g(x) = \frac{1}{x-2} (domain \mathbb{R} \setminus \{2\}). Then:

  • Domain of f + g: \mathbb{R} \setminus \{0\} \cap \mathbb{R} \setminus \{2\} = \mathbb{R} \setminus \{0, 2\}.
  • Domain of \frac{f}{g}: start with \mathbb{R} \setminus \{0, 2\}, then exclude x where g(x) = 0. But \frac{1}{x-2} is never zero (a reciprocal is never zero), so no additional exclusions. Domain: \mathbb{R} \setminus \{0, 2\}.

The rule is mechanical: intersect the domains, then (for quotients) exclude the zeros of the denominator function.

Seeing the operations on a graph

When you add two functions, the y-value of the sum at each x is the stack of the two individual y-values. Here is what f(x) = x and g(x) = \sin x look like individually, and what their sum h(x) = x + \sin x looks like.

Graphs of f(x) = x, g(x) = sin(x), and their sum f+gThree curves on the same axes. A straight line y = x in muted colour. A sine wave in muted colour. And the sum y = x + sin x in bold, which looks like a wavy line oscillating around y = x. x y π 2π −π y = x sin x x + sin x
The dashed lines show $y = x$ and $y = \sin x$ individually. The bold red curve is their sum $y = x + \sin x$ — at every $x$, add the two heights. The result is a line that wobbles, oscillating around $y = x$ with amplitude $1$.

For the product f(x) \cdot g(x) = x \sin x, the picture is different: the sine wave's amplitude grows as |x| grows, because the multiplier x scales the wave up.

Properties

The arithmetic of functions inherits properties from the arithmetic of real numbers, because at each x you are just combining two real numbers.

Commutativity.

f + g = g + f, \qquad f \cdot g = g \cdot f

Addition and multiplication of functions are commutative. Subtraction and division are not: f - g \ne g - f in general, and \frac{f}{g} \ne \frac{g}{f} in general.

Associativity.

(f + g) + h = f + (g + h), \qquad (f \cdot g) \cdot h = f \cdot (g \cdot h)

Distributivity.

f \cdot (g + h) = f \cdot g + f \cdot h

Identity elements. The zero function \mathbf{0}(x) = 0 is the additive identity: f + \mathbf{0} = f. The constant function \mathbf{1}(x) = 1 is the multiplicative identity: f \cdot \mathbf{1} = f.

Scalar multiplication distributes.

c(f + g) = cf + cg, \qquad (c_1 + c_2)f = c_1 f + c_2 f

All of these hold on the common domain. The properties are saying: for every x in the intersection of all relevant domains, the real-number identities hold.

Summary of algebraic properties of function operationsA table-like diagram showing four properties: commutativity for addition and multiplication, associativity, distributivity, and identity elements. Commutative f + g = g + f f · g = g · f Associative (f + g) + h = f + (g + h) (f · g) · h = f · (g · h) Distributive f · (g + h) = f · g + f · h Identities f + 0 = f f · 1 = f
The four key properties of function arithmetic — all inherited from the corresponding properties of real-number arithmetic.

Two worked examples

Example 1: Given $f(x) = x^2 + 1$ and $g(x) = \sqrt{x}$, find $(f + g)(x)$, $(f \cdot g)(x)$, and $\left(\frac{f}{g}\right)(x)$, with domains

Step 1. Determine the individual domains. f(x) = x^2 + 1 is a polynomial — defined for all x \in \mathbb{R}. g(x) = \sqrt{x} requires x \ge 0, so D_g = [0, \infty).

Why: always find the individual domains first, because the combined domain depends on both.

Step 2. Find the common domain. D_f \cap D_g = \mathbb{R} \cap [0, \infty) = [0, \infty).

Why: both functions must be defined at the same x for the combination to make sense.

Step 3. Write each combined function.

(f + g)(x) = x^2 + 1 + \sqrt{x}, \quad \text{domain } [0, \infty)
(f \cdot g)(x) = (x^2 + 1)\sqrt{x} = x^{5/2} + x^{1/2}, \quad \text{domain } [0, \infty)

Why: distribute the multiplication and simplify using x^2 \cdot x^{1/2} = x^{5/2}.

Step 4. For the quotient, add the extra check. g(x) = \sqrt{x} = 0 when x = 0. So exclude 0.

\left(\frac{f}{g}\right)(x) = \frac{x^2 + 1}{\sqrt{x}}, \quad \text{domain } (0, \infty)

Why: x = 0 makes g(x) = 0, and dividing by zero is undefined. The bracket changes from [0 to (0.

Result: The sum and product have domain [0, \infty). The quotient has domain (0, \infty) — a subtle but important difference: the left endpoint is excluded.

Graphs of f, g, and f+g for Example 1Three curves on the same axes for x from 0 to 3. The parabola y equals x squared plus 1 rises steeply. The square root curve y equals root x rises gently. Their sum y equals x squared plus 1 plus root x is drawn in red, above both. x y 1 2 1 2 5 f(x) g(x) (f+g)(x) (0, 1)
The dashed curves show $f(x) = x^2 + 1$ and $g(x) = \sqrt{x}$ individually. The solid red curve is their sum $(f+g)(x) = x^2 + 1 + \sqrt{x}$. At each $x$, the red curve's height is the sum of the two dashed heights. At $x = 0$, $f(0) + g(0) = 1 + 0 = 1$.

The graph confirms the algebra. At x = 0, the sum equals 1 (the parabola contributes 1, the square root contributes 0). As x grows, the x^2 term dominates and the sum curve climbs steeply.

Example 2: Given $f(x) = \frac{1}{x+1}$ and $g(x) = \frac{1}{x-1}$, find $(f - g)(x)$ and $\left(\frac{f}{g}\right)(x)$, with domains

Step 1. Find the individual domains. D_f = \mathbb{R} \setminus \{-1\} (exclude x = -1). D_g = \mathbb{R} \setminus \{1\} (exclude x = 1).

Why: each function has a denominator that must not be zero.

Step 2. Intersect the domains. D_f \cap D_g = \mathbb{R} \setminus \{-1, 1\}.

Why: the combined function needs both f and g to be defined, so exclude the problem points of both.

Step 3. Compute the difference.

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

Why: combine the fractions over a common denominator (x+1)(x-1) = x^2 - 1.

Step 4. Compute the quotient. For \frac{f}{g}, first note that g(x) = 0 would require \frac{1}{x-1} = 0, which has no solution (a reciprocal is never zero). So no extra exclusions.

\left(\frac{f}{g}\right)(x) = \frac{1/(x+1)}{1/(x-1)} = \frac{x-1}{x+1}

Domain: \mathbb{R} \setminus \{-1, 1\}.

Why: dividing by \frac{1}{x-1} is the same as multiplying by (x-1). No additional zeros to exclude.

Result: (f - g)(x) = \frac{-2}{x^2 - 1} on \mathbb{R} \setminus \{-1, 1\}. \left(\frac{f}{g}\right)(x) = \frac{x-1}{x+1} on \mathbb{R} \setminus \{-1, 1\}.

Graph of f/g = (x-1)/(x+1) showing asymptotesThe graph of y equals (x minus 1) over (x plus 1). A vertical asymptote at x equals negative 1 and a horizontal asymptote at y equals 1. The curve passes through the origin at (0, negative 1) and through (1, 0). x y −1 1 2 −3 1 −1 y = 1 x = −1 (0, −1) (1, 0)
The graph of $\frac{f}{g}(x) = \frac{x-1}{x+1}$. The vertical asymptote at $x = -1$ reflects the excluded domain point. The horizontal asymptote at $y = 1$ shows the long-run behaviour. The curve passes through $(0, -1)$ and $(1, 0)$.

The simplified form \frac{x-1}{x+1} looks like a new function defined on \mathbb{R} \setminus \{-1\}. But as f/g, its domain must also exclude x = 1 (where g is undefined), even though the simplified formula has no problem at x = 1. This is a common subtlety: simplification can hide domain restrictions that the original form reveals.

Building functions from pieces

The operations above are not just abstract — they are how many standard functions are built. A polynomial p(x) = a_n x^n + \cdots + a_1 x + a_0 is a sum of scaled power functions. A rational function \frac{p(x)}{q(x)} is a quotient of two polynomials. Understanding function arithmetic is understanding how complicated functions are assembled from simple ones.

Here is a useful pattern: if f and g are both polynomials (defined on all of \mathbb{R}), then f + g, f - g, and f \cdot g are also polynomials (defined on all of \mathbb{R}). The quotient f/g is a rational function, defined everywhere except at the roots of g.

How polynomials and rational functions are builtA flow diagram. Constant functions and the identity function x combine via sum and product to form polynomials. Dividing polynomials gives rational functions. constants x +, · polynomials ÷ rational functions
Polynomials are built from constants and $x$ using addition and multiplication. Rational functions are built by dividing one polynomial by another. Each step is an instance of the algebra of functions.

Interactive: combine two functions

Choose an operation, slide $x$, and watch three graphs: $f(x) = x$ (dashed), $g(x) = x^2$ (dashed), and their combination (solid red). The vertical bars at the current $x$ add, subtract, multiply, or divide pointwise \u2014 the solid curve is the continuous record of those pointwise combinations. Press Play to sweep $x$ across the axis and watch the combined curve being traced.

Common confusions

  • "The domain of f + g is D_f \cup D_g." It is the intersection, not the union. Both functions must be defined at the same x. If f works at x = 5 but g does not, you cannot compute f(5) + g(5).

  • "After simplifying \frac{f}{g}, the domain expands." Simplifying a formula does not change the domain. If g(x) = 0 at some x, that x is excluded from the domain of f/g regardless of what the simplified expression looks like.

  • "The product of two functions is always harder to find the domain of." The domain of f \cdot g is the same as the domain of f + g — just D_f \cap D_g. Multiplication does not introduce any extra restrictions.

  • "f \cdot g = 0 means f = 0 or g = 0 as functions." No. f \cdot g can be zero at specific x-values without either function being identically zero. (f \cdot g)(x) = 0 means f(x) = 0 or g(x) = 0 at that particular x.

  • "Scalar multiplication cf changes the domain." It does not. Multiplying every output by a constant does not affect which inputs are allowed. D_{cf} = D_f for any constant c.

The material above covers all the function-operations content in the class 11 syllabus. The following extends the ideas for those who want a more structural view.

The set of all real-valued functions forms a ring

If you restrict to functions defined on a common domain D, the set \mathcal{F}(D, \mathbb{R}) of all functions from D to \mathbb{R} has a rich algebraic structure. Under addition and multiplication:

  • (f + g)(x) = f(x) + g(x) makes \mathcal{F} an abelian group under + (with the zero function as identity, and -f as the additive inverse).
  • (f \cdot g)(x) = f(x) \cdot g(x) makes \mathcal{F} a commutative ring with unity (the constant function 1).

This means every property of ring arithmetic — distributivity, associativity, existence of identities — holds automatically. The algebraic structure of functions is exactly the algebraic structure of real numbers, lifted pointwise.

Domain subtleties with rational functions

When you write h(x) = \frac{x^2 - 1}{x - 1}, you might be tempted to cancel: h(x) = \frac{(x-1)(x+1)}{x-1} = x + 1. But the original formula is undefined at x = 1, while x + 1 is defined everywhere. The correct statement is:

h(x) = x + 1 \quad \text{for all } x \ne 1

The simplified function x + 1 and the original \frac{x^2-1}{x-1} agree everywhere except at x = 1, where the original has a "hole." In the language of limits, you would say the function has a removable discontinuity at x = 1.

This subtlety matters when you combine functions: the domain of the result is determined by the original formulas, not by any simplified version.

Boundedness under operations

If f and g are both bounded on a domain D, what can you say about their combinations?

  • f + g is bounded. If |f(x)| \le M_1 and |g(x)| \le M_2, then |f(x) + g(x)| \le M_1 + M_2.
  • f \cdot g is bounded. |f(x) \cdot g(x)| \le M_1 \cdot M_2.
  • f/g may be unbounded. Even if g is bounded, it can approach zero, making the quotient large. For instance, f(x) = 1 and g(x) = x on (0, 1] are both bounded, but f/g = 1/x is unbounded.

The quotient is the operation that can break boundedness — and this connects to why division by zero is excluded from the domain.

Where this leads next

You now know how to combine functions using the four arithmetic operations, how to determine the domain of each combination, and what algebraic properties these operations satisfy. Here is where this connects.

  • Composite Functions — instead of combining outputs at the same input, feed one function's output into another function's input. This is a fundamentally different way to combine functions.
  • Functions — Definition and Notation — the foundational definitions that underpin the domain rules here.
  • Domain and Range — how domain restrictions arise and how to find them for individual functions.
  • Types of Functions — injective, surjective, bijective — classifications that interact with function operations.
  • Operations and Properties — the real-number properties (commutativity, associativity, distributivity) that function arithmetic inherits.