Introduction Of Functions

Q: What is the difference between a function and a relation?

A relation from X to Y is any set of ordered pairs (x,y). A function is a special relation where every element of X appears in exactly one pair - each input has exactly one output. Every function is a relation, but not every relation is a function.

Q: What is the difference between range and co-domain?

Co-domain is the target set Y declared when defining f:X to Y. Range is the actual set of outputs {f(x):x in X}. Range is always a subset of co-domain; they are equal only when f is onto (surjective).

Q: How do I find the domain of a function quickly?

Start with all real numbers and remove: points where the denominator is zero; points where an even root has a negative argument; points where a log has a non-positive argument or invalid base; and points outside the natural domain of inverse trig functions.

Q: What is the vertical line test?

A graphical check: if every vertical line intersects the graph at most once, the graph represents a function. If some vertical line cuts the graph at two or more points, one input maps to multiple outputs, so it is not a function.

Q: Can a function have a domain that is not all real numbers?

Yes. The domain is whatever set of inputs makes the rule valid. For sqrt(x) it is [0, infinity); for 1/x it is R minus {0}; for arcsin(x) it is [-1, 1].

Q: How is the range of f/g different from the domain of f/g?

Domain of f/g is where both f and g are defined and g(x) is not zero. Range of f/g is the set of actual output values, which usually needs a separate calculation.

Q: Is f(x)=x a function?

Yes. The identity function f(x)=x maps every real number to itself. Its domain and range are both R, and it trivially satisfies the vertical line test.

MathsFunctionsFor JEE aspirants

A function from set $X$ to set $Y$ is a rule that assigns to each element of $X$ exactly one element of $Y$ . Written as $f : X \to Y$ with $y = f (x)$ , where $x$ is the pre-image and $y$ is the image. Every element of $X$ must have a unique image; some elements of $Y$ may have no pre-image. The set $X$ is the domain, $Y$ is the co-domain, and the collection of all images $f (x)$ is the range. This concept covers definition, value of a function, algebra of functions, and how to find domain and range.

Key Formulas - Quick Reference

Function: $f : X \to Y$ , $y = f (x)$ with $x \in X$ , $y \in Y$
Range: $f (X) = {f (x) : x \in X} \subseteq Y$ (co-domain)
Sum / difference / product: $(f \pm g) (x) = f (x) \pm g (x)$ , $(f \cdot g) (x) = f (x) g (x)$ , domain $= D_{1} \cap D_{2}$
Quotient: $(f / g) (x) = \frac{f ( x )}{g ( x )}$ , domain $= (D_{1} \cap D_{2}) ∖ {x : g (x) = 0}$
Domain rules: denominator $\neq = 0$ ; expression under even root $\geq 0$ ; argument of $lo g$ must be $> 0$ ; base of $lo g$ must be $> 0$ and $\neq = 1$
Domain of inverse trig: $sin^{- 1} x, cos^{- 1} x$ need $- 1 \leq x \leq 1$ ; $tan^{- 1} x, cot^{- 1} x$ for all real $x$ ; $sec^{- 1} x, csc^{- 1} x$ need $∣ x ∣ \geq 1$

1. Definition of a Function

Let $X$ and $Y$ be any two non-empty sets. A function from $X$ to $Y$ is a rule of correspondence that assigns to each element of $X$ one and only one element of $Y$ . If the rule is called $f$ , we write $f : X \to Y$ where $y = f (x)$ , with $x \in X$ and $y \in Y$ . Here $y$ is the image of $x$ under $f$ , and $x$ is the pre-image of $y$ .

A mapping $f : X \to Y$ is a function if every element of $X$ has an image in $Y$ . Some elements of $Y$ may not be the image of anything in $X$ - that is allowed.
Each element of $X$ has one and only one image. A function cannot be multi-valued. A multi-valued mapping is called a relation, not a function.

Figure 1: A function maps each element of X to exactly one element of Y

Graphical (Vertical Line) Test

To check whether a relation between $x$ and $y$ is a function, draw a line parallel to the $y$ -axis. If every such vertical line meets the graph at one and only one point, the relation is a function. If any vertical line meets the graph at two or more points, it is not a function.

Figure 2: The vertical line test - a valid function is cut by any vertical line at exactly one point

2. Value of a Function

The value of a function $y = f (x)$ at $x = a$ is written $f (a)$ and is obtained by substituting $x = a$ into the expression for $f (x)$ .

Solved Example 1

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

for

x \in (- 1, 1)

, find

f (0)

and

f (1/2)

Solution:

$f (0) = lo g (\frac{1 - 0}{1 + 0}) = lo g 1 = 0$ .

$f (1/2) = lo g (\frac{1 - 1/2}{1 + 1/2}) = lo g \frac{1/2}{3/2} = lo g \frac{1}{3} = - lo g 3$ .

3. Algebra of Functions

Let $f : D_{1} \to R$ and $g : D_{2} \to R$ be two real-valued functions. We can build new functions from them.

Sum: $(f + g) (x) = f (x) + g (x)$ , defined on $D = D_{1} \cap D_{2}$
Difference: $(f - g) (x) = f (x) - g (x)$ , defined on $D = D_{1} \cap D_{2}$
Product: $(f \cdot g) (x) = f (x) g (x)$ , defined on $D = D_{1} \cap D_{2}$
Quotient: $(f / g) (x) = \frac{f ( x )}{g ( x )}$ , defined on $D = (D_{1} \cap D_{2}) ∖ {x : g (x) = 0}$

Solved Example 2

Let

f (x) = sin^{- 1} x

and

g (x) = (3 - x) (x + 1)

. Find

f + g

f - g

f \cdot g

, and

f / g

Solution:

Domain of $f$ : $- 1 \leq x \leq 1$ . Domain of $g$ : $(3 - x) (x + 1) \geq 0 \Rightarrow - 1 \leq x \leq 3$ . Common domain $D_{1} \cap D_{2} = [- 1, 1]$ .

$(f + g) (x) = sin^{- 1} x + (3 - x) (1 + x)$ , defined on $[- 1, 1]$ .

$(f - g) (x) = sin^{- 1} x - (3 - x) (1 + x)$ , defined on $[- 1, 1]$ .

$(f \cdot g) (x) = sin^{- 1} x \cdot (3 - x) (1 + x)$ , defined on $[- 1, 1]$ .

$(f / g) (x) = \frac{sin ^{- 1} x}{( 3 - x ) ( x + 1 )}$ , defined on $(- 1, 1]$ (excluding $x = - 1$ where $g = 0$ ).

Solved Example 3

Solve for

x

lo g_{1/4} x \geq 1

Solution:

Rewrite $1$ as $lo g_{1/4} (1/4)$ : $lo g_{1/4} x \geq lo g_{1/4} (1/4)$ .

Since the base $1/4$ lies between $0$ and $1$ , the log is a decreasing function, so the inequality flips:

$x \leq \frac{1}{4}$ .

Also $lo g x$ needs $x > 0$ . Combining: $x \in (0, \frac{1}{4}]$ .

4. Domain and Range of a Function

For $f : X \to Y$ :

Domain $= X$ (the set of inputs).
Co-domain $= Y$ (the target set).
Range $= f (X) = {f (x) : x \in X}$ , the set of actual outputs. Range is always a subset of the co-domain: $f (X) \subseteq Y$ .

Common Domain Rules

Denominator must be non-zero.
Expression under an even root (, fourth root, ...) must be $\geq 0$ .
Argument of $lo g$ must be $> 0$ ; base of $lo g$ must be $> 0$ and $\neq = 1$ .
For $sin^{- 1} x$ and $cos^{- 1} x$ : $- 1 \leq x \leq 1$ .
For $tan^{- 1} x$ and $cot^{- 1} x$ : all real $x$ .
For $sec^{- 1} x$ and $csc^{- 1} x$ : $∣ x ∣ \geq 1$ .

Problems Based on Domain

Solved Example 4

Find the domain of

f (x) = \frac{x ^{2} + 1}{x ^{2} - 4}

Solution:

The denominator must be non-zero: $x^{2} - 4 \neq = 0 \Rightarrow x \neq = \pm 2$ .

Domain $= R - {- 2, 2}$ .

Solved Example 5

Find the domain of

f (x) = (\frac{x}{1 - ∣ x ∣})^{1/2000}

Solution:

Since $1/2000$ is an even-denominator rational exponent, we need $\frac{x}{1 - ∣ x ∣} \geq 0$ .

Case 1: $x \geq 0$ , so $∣ x ∣ = x$ and the expression becomes $\frac{x}{1 - x}$ . We need $\frac{x}{1 - x} \geq 0$ , which gives $x \in [0, 1)$ .

Case 2: $x < 0$ , so $∣ x ∣ = - x$ and the expression becomes $\frac{x}{1 + x}$ . We need $\frac{x}{1 + x} \geq 0$ , which gives $x \in (- \infty, - 1) \cup [0, \infty)$ . But we are in Case 2 ( $x < 0$ ), so we keep $x \in (- \infty, - 1)$ .

Taking the union: Domain $= (- \infty, - 1) \cup [0, 1)$ .

Solved Example 6

Find the domain of

lo g_{x + 3} (x^{2} + 2 x + 1)

Solution:

Two conditions:

(i) Argument $> 0$ : $x^{2} + 2 x + 1 > 0 \Rightarrow (x + 1)^{2} > 0 \Rightarrow x \neq = - 1$ .

(ii) Base valid: $x + 3 > 0$ and $x + 3 \neq = 1 \Rightarrow x > - 3$ and $x \neq = - 2$ .

Domain $= (- 3, \infty) - {- 2, - 1}$ .

Solved Example 7

Find the domain of

cos^{- 1} (\frac{6 - 3 x}{4}) + csc^{- 1} (\frac{x - 1}{2})

Solution:

For $cos^{- 1} (\frac{6 - 3 x}{4})$ we need $- 1 \leq \frac{6 - 3 x}{4} \leq 1$ .

Multiplying by 4: $- 4 \leq 6 - 3 x \leq 4 \Rightarrow \frac{2}{3} \leq x \leq \frac{10}{3}$ ... (1)

For $csc^{- 1} (\frac{x - 1}{2})$ we need $\frac{x - 1}{2} \geq 1$ , i.e. $\frac{x - 1}{2} \leq - 1$ or $\frac{x - 1}{2} \geq 1$ .

So $x \leq - 1$ or $x \geq 3$ , giving $x \in (- \infty, - 1] \cup [3, \infty)$ ... (2)

Intersecting (1) and (2): only $x = 3$ lies in both.

Domain $= {3}$ .

Problems Based on Range

Solved Example 8

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

, find the range of

f

Solution:

Let $y = \frac{x ^{2} + 2 x + 3}{x}$ . Cross-multiplying: $x^{2} + (2 - y) x + 3 = 0$ .

Since $x$ is real, the discriminant must be $\geq 0$ :

$(2 - y)^{2} - 4 (1) (3) \geq 0 \Rightarrow (2 - y)^{2} \geq 12 \Rightarrow ∣2 - y ∣ \geq 23$ .

So $y \leq 2 - 23$ or $y \geq 2 + 23$ .

Range $= (- \infty, 2 - 23] \cup [2 + 23, \infty)$ .

Solved Example 9

Find the range of

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

Solution:

Let $y = \frac{x ^{2}}{x ^{2} + 1}$ . Solving for $x^{2}$ : $y (x^{2} + 1) = x^{2} \Rightarrow x^{2} (1 - y) = y \Rightarrow x^{2} = \frac{y}{1 - y}$ .

Since $x^{2} \geq 0$ , we need $\frac{y}{1 - y} \geq 0$ , which gives $y \in [0, 1)$ .

Range $= [0, 1)$ .

Alternate method (substitution): Put $x = tan θ$ where $θ \in (- π /2, π /2)$ . Then $y = \frac{tan ^{2} θ}{sec ^{2} θ} = sin^{2} θ \in [0, 1)$ .

Solved Example 10

Find the range of

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 1

for

x \in [- 1, 3]

Solution:

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ .

$f$ is increasing on $(- \infty, 1]$ and $[2, \infty)$ , decreasing on $[1, 2]$ .

Evaluate at critical points and endpoints:

$f (- 1) = - 2 - 9 - 12 + 1 = - 22$ ; $f (1) = 2 - 9 + 12 + 1 = 6$ ; $f (2) = 16 - 36 + 24 + 1 = 5$ ; $f (3) = 54 - 81 + 36 + 1 = 10$ .

Minimum $= - 22$ , maximum $= 10$ .

Range $= [- 22, 10]$ .

Solved Example 11

Find the range of

sin^{- 1} x + cos^{- 1} x + tan^{- 1} x

Solution:

The domain is $[- 1, 1]$ (from $sin^{- 1}$ and $cos^{- 1}$ ).

We know $sin^{- 1} x + cos^{- 1} x = \frac{π}{2}$ for all $x \in [- 1, 1]$ .

So $f (x) = \frac{π}{2} + tan^{- 1} x$ .

On $[- 1, 1]$ , $tan^{- 1} x \in [- \frac{π}{4}, \frac{π}{4}]$ .

Range $= [\frac{π}{2} - \frac{π}{4}, \frac{π}{2} + \frac{π}{4}] = [\frac{π}{4}, \frac{3 π}{4}]$ .

Common Mistakes to Avoid

Watch out

Forgetting to flip the inequality when the log base lies between $0$ and $1$ . Base $< 1$ means log is decreasing.
Cancelling common terms without checking the domain of that term. For example, cancelling $\frac{1}{x - 2.7}$ from both sides is only valid when $x \neq = 2.7$ .
Using $x = sin θ$ instead of $x = tan θ$ when substituting - this artificially restricts the domain.
Assuming range of $tan^{- 1} x$ is always $(- \frac{π}{2}, \frac{π}{2})$ - it is only that full interval when $x$ ranges over all of $R$ . If $x$ is restricted, $tan^{- 1} x$ is also restricted.
Forgetting that division by zero excludes points from the domain (even after algebraic simplification).

Frequently Asked Questions

Q1. What is the difference between a function and a relation?

A relation from $X$ to $Y$ is any set of ordered pairs $(x, y)$ . A function is a special relation where every element of $X$ appears in exactly one pair - each input has exactly one output. So every function is a relation, but not every relation is a function.

Q2. What is the difference between range and co-domain?

The co-domain is the target set $Y$ declared when defining $f : X \to Y$ . The range is the actual set of outputs ${f (x) : x \in X}$ . Range is always a subset of co-domain; they are equal only when $f$ is onto (surjective).

Q3. How do I find the domain of a function quickly?

Start with all real numbers, then remove: (i) points where the denominator is zero, (ii) points where an even root has a negative argument, (iii) points where a log has a non-positive argument or invalid base, (iv) points outside the natural domain of inverse trig functions.

Q4. What is the vertical line test?

It is a quick graphical check: if every vertical line ( $x =$ constant) intersects the graph at most once, the graph represents a function. If some vertical line cuts the graph at two or more points, the graph is not a function because one input would map to multiple outputs.

Q5. Can a function have a domain that is not all real numbers?

Yes. The domain is whatever set of inputs makes the rule valid. For $f (x) = x$ the natural domain is $[0, \infty)$ ; for $f (x) = 1/ x$ it is $R - {0}$ ; for $sin^{- 1} x$ it is $[- 1, 1]$ .

Q6. How is the range of $f / g$ different from the domain of $f / g$ ?

The domain of $f / g$ is where both $f$ and $g$ are defined and $g (x) \neq = 0$ . The range of $f / g$ is the actual set of output values ${f (x) / g (x)}$ as $x$ varies over that domain, which usually needs a separate calculation.

Q7. Is $f (x) = x$ a function?

Yes. The identity function $f (x) = x$ maps every real number to itself. Its domain and range are both $R$ , and it satisfies the vertical line test trivially.

Ready to master Functions?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. Definition of a Function

Graphical (Vertical Line) Test

2. Value of a Function

3. Algebra of Functions

4. Domain and Range of a Function

Common Domain Rules

Problems Based on Domain

Problems Based on Range

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between a function and a relation?

Q2. What is the difference between range and co-domain?

Q3. How do I find the domain of a function quickly?

Q4. What is the vertical line test?

Q5. Can a function have a domain that is not all real numbers?

Q6. How is the range of f/g different from the domain of f/g?

Q7. Is f(x)=x a function?

Ready to master Functions?

Q6. How is the range of $f / g$ different from the domain of $f / g$ ?

Q7. Is $f (x) = x$ a function?