Functions and their Graphs

What Makes a Relation a Function?

Among all relations, one type is so important that the rest of mathematics is essentially built on it.

Key Point (Definition): A relation f from a set A to a set B is a function if every element of A has one and only one image in B. We write $f : A \to B$, and if $(a, b) \in f$ we write $f(a) = b$ — b is the image of a, and a is a preimage of b.

Unpack the definition into the two conditions you actually test:

1. Every element of A is used — the domain of f is all of A; no input may be left without an image.
2. No input has two images — no two distinct ordered pairs share the same first element.

A function is also called a map or mapping — picture it as a machine: feed in any element of A, exactly one element of B comes out. Feed in the same input twice, get the same output twice.

The test in action

• R = {(2,1), (3,1), (4,2)} from {2, 3, 4}: every input appears once — function. (Two inputs sharing an output is fine!)
• R = {(2,2), (2,4), (3,3), (4,4)}: the input 2 has two images (2 and 4) — not a function.
• R = {(1,2), (2,3), (3,4), (4,5), (5,6), (6,7)}: each input once — function.
• R = {(x, y) : y = x + 1} on A = {1, …, 6} (Section 2's arrow diagram): the element 6 has NO image — not a function from A to A.

[Board Important] "Same output, different inputs" never disqualifies a function; "same input, different outputs" always does. On an arrow diagram: every domain point fires exactly one arrow. On a graph: every vertical line meets the curve at most once (the vertical line test).

Real Functions, and the First Four Standard Graphs

Key Point (Definition): A function whose range is $\mathbb{R}$ or a subset of $\mathbb{R}$ is a real valued function; if its domain is also $\mathbb{R}$ or a subset of $\mathbb{R}$, it is a real function.

From here on, our functions are real functions given by formulas like $f(x) = 2x + 1$, and each deserves a mental picture — its graph, the set of points $(x, f(x))$.

(i) Identity function: $f(x) = x$

Every input maps to itself. Domain $\mathbb{R}$, range $\mathbb{R}$. The graph is the straight line through the origin at 45°.

(ii) Constant function: $f(x) = c$

Every input maps to the same value c. Domain $\mathbb{R}$, range {c}. The graph is a horizontal line.

(iii) Polynomial functions: $f(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$

Here n is a non-negative integer and the coefficients are real. Examples: $f(x) = x^3 - x^2 + 2$. Non-examples: $h(x) = x^{2/3} + 2x$ — the exponent $\frac{2}{3}$ is not a non-negative integer.

The two model graphs: $y = x^2$ (a parabola through the origin, range $[0, \infty)$) and $y = x^3$ (rising through the origin, range $\mathbb{R}$).

(iv) Rational functions: $f(x) = \frac{f_1(x)}{g_1(x)}$, quotients of polynomials, defined where $g_1(x) \neq 0$

The model: $f(x) = \frac{1}{x}$ with domain $\mathbb{R} - \{0\}$ and range $\mathbb{R} - \{0\}$ — the two-branch hyperbola.

[JEE Tip] Memorise each standard function as a domain-range pair — half the MCQs on this section reduce to quoting one. $x^2$: ($\mathbb{R}$, $[0, \infty)$); $x^3$: ($\mathbb{R}$, $\mathbb{R}$); $\frac{1}{x}$: ($\mathbb{R} - \{0\}$, $\mathbb{R} - \{0\}$).

Modulus, Signum and Greatest Integer

(v) Modulus function: $f(x) = |x|$

$$f(x) = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$$

For negative x, the output is the negative of x — hence positive. Domain $\mathbb{R}$, range $[0, \infty)$. The graph is the V with its corner at the origin.

(vi) Signum function

$$f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$$

It reports only the sign of the input. Domain $\mathbb{R}$, range {$-1$, 0, 1}. For $x \neq 0$ it equals $\frac{|x|}{x}$ (or $\frac{x}{|x|}$).

(vii) Greatest integer function: $f(x) = [x]$

$[x]$ is the greatest integer less than or equal to x:

• $[2.7] = 2$, $[0.4] = 0$, $[3] = 3$
• $[-0.5] = -1$ and $[-2.1] = -3$ — for negatives, round DOWN, not toward zero!

On each interval $[n, n + 1)$ the value is the constant n, producing the staircase graph. Domain $\mathbb{R}$, range $\mathbb{Z}$.

Key Point: $n \leq x < n + 1 \iff [x] = n$. This equivalence is how every $[x]$ equation and inequality is solved.

[JEE Tip] The single most-tested value type: $[-2.1] = -3$ (not $-2$). If this feels wrong, recall the definition — the greatest integer NOT exceeding $-2.1$ sits below it on the number line.

Solved Examples

Example 1. Examine each relation and state whether it is a function: (i) R = {(2,1), (3,1), (4,2)} (ii) R = {(2,2), (2,4), (3,3), (4,4)} (iii) R = {(1,2), (2,3), (3,4), (4,5), (5,6), (6,7)}.

Solution.

Step 1 — recall the test. Scan the FIRST coordinates: a relation is a function exactly when no first coordinate repeats with different partners (and every domain element appears).

(i) Step 2. First coordinates 2, 3, 4 — each appears once. The fact that 2 and 3 SHARE the image 1 is irrelevant (same output is always allowed). Function.

(ii) Step 3. The first coordinate 2 appears twice, with images 2 and 4 — one input, two outputs. Not a function.

(iii) Step 4. First coordinates 1 through 6, each exactly once. Function.

Takeaway: \"same output, different inputs\" never disqualifies; \"same input, different outputs\" always does.

Example 2. R is defined on $\mathbb{N}$ by R = {(x, y) : y = 2x, x, y $\in \mathbb{N}$}. What are its domain, codomain and range? Is it a function?

Solution.

Step 1 — check the domain. For every natural x, the image $2x$ is again a natural number — no input is ever dropped: domain = $\mathbb{N}$.

Step 2 — codomain and range. Codomain = $\mathbb{N}$ (the announced target). The images are $2, 4, 6, \ldots$ — the even naturals: range = {2, 4, 6, …}.

Step 3 — function test. Each x determines the single value 2x — one image per input. Yes, a function.

Takeaway: range $\neq$ codomain is perfectly fine for a function — the odd naturals are simply never hit. (Whether range = codomain is what \"onto\" measures in Class 12.)

Example 3. $f : \mathbb{N} \to \mathbb{N}$ is defined by $f(x) = 2x + 1$. Complete the table for x = 1 to 7.

Solution.

Step 1 — substitute each value. $f(1) = 2(1) + 1 = 3$; $f(2) = 5$; $f(3) = 7$; $f(4) = 9$; $f(5) = 11$; $f(6) = 13$; $f(7) = 15$.

Step 2 — observe the pattern. The outputs are consecutive odd numbers from 3 — each step in x adds 2 to the output, as the coefficient of x dictates.

Takeaway: a formula-defined function is evaluated by pure substitution; the coefficient of x controls the stride of the output list.

Example 4. A function is defined by $f(x) = 2x - 5$. Find (i) f(0) (ii) f(7) (iii) f($-3$).

Solution.

(i) Step 1. $f(0) = 2(0) - 5 = 0 - 5 = -5$.

(ii) Step 2. $f(7) = 2(7) - 5 = 14 - 5 = 9$.

(iii) Step 3 — signs intact. $f(-3) = 2(-3) - 5 = -6 - 5 = -11$. (Not $-6 + 5 = -1$ — subtracting 5 pushes further negative.)

Takeaway: substitute with brackets around negative inputs; the double negative in (iii) is where marks are lost.

Example 5. The function t maps Celsius to Fahrenheit: $t(C) = \frac{9C}{5} + 32$. Find (i) t(0) (ii) t(28) (iii) t($-10$) (iv) the value of C when t(C) = 212.

Solution.

(i) Step 1. $t(0) = \frac{0}{5} + 32 = 32$ — freezing point: 0°C = 32°F.

(ii) Step 2. $t(28) = \frac{9 \times 28}{5} + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4$.

(iii) Step 3. $t(-10) = \frac{9 \times (-10)}{5} + 32 = \frac{-90}{5} + 32 = -18 + 32 = 14$.

(iv) Step 4 — reverse the function. Set the output: $\frac{9C}{5} + 32 = 212$. Subtract 32: $\frac{9C}{5} = 180$. Multiply by $\frac{5}{9}$: $C = \frac{180 \times 5}{9} = 100$.

Step 5 — physical check. Water boils at 100°C = 212°F ✓ — the algebra matches the physics.

Takeaway: part (iv) runs the machine backwards: fix the output, solve for the input — the first taste of inverse functions.

Example 6. Evaluate: (i) $|{-7}|$ (ii) $|3 - 8|$ (iii) $f(-4)$ for $f(x) = |x - 1|$. Where is the corner of the graph of $|x - 1|$?

Solution.

(i) Step 1. $|-7| = 7$ — modulus strips the sign.

(ii) Step 2 — inside first. $3 - 8 = -5$, then $|-5| = 5$.

(iii) Step 3. $f(-4) = |-4 - 1| = |-5| = 5$.

Step 4 — locate the corner. The V-corner sits where the inside vanishes: $x - 1 = 0$, i.e. x = 1. The graph of $|x - 1|$ is the V of $|x|$ shifted RIGHT by 1.

Takeaway: always work inside-out — evaluate the inside, then strip the sign; corners of modulus graphs sit at the zeros of the inside.

Example 7. For the signum function, find f(7), f($-4.2$), f(0), and the range of f over the domain $[-3, -1]$.

Solution.

Step 1 — apply the three-case definition. $f(7) = 1$ (input positive); $f(-4.2) = -1$ (negative); $f(0) = 0$ (the special middle case).

Step 2 — restrict to $[-3, -1]$. Every number in this interval is negative, so the signum is constantly $-1$ there.

Step 3 — range over the restriction. Only one value occurs: range = {$-1$}.

Takeaway: signum ignores magnitude entirely — restricting the domain to one sign collapses the range to a single value.

Example 8. Evaluate: (i) 3.7 $-1.2$ 5 $-0.01$ [1.99].

Solution.

Step 1 — recall the definition. [x] = the greatest integer LESS THAN OR EQUAL to x — on the number line, drop to the nearest integer at or below x.

(i) Step 2. $[3.7] = 3$ (truncate the positive).

(ii) Step 3. $[-1.2]$: the integers at or below $-1.2$ are $-2, -3, \ldots$; the greatest is $-2$. (NOT $-1$, which exceeds $-1.2$.)

(iii) Step 4. $[5] = 5$ — integers are their own floor.

(iv) Step 5. $[-0.01] = -1$ — even a hair below zero drops to $-1$.

(v) Step 6. $[1.99] = 1$ — still short of 2, no matter how close.

Takeaway: positives truncate; negatives drop DOWN — the pair $[-1.2] = -2$ and $[1.99] = 1$ covers both classic traps.

Example 9. Find the domain and range of $f(x) = \sqrt{9 - x^2}$.

Solution.

Step 1 — domain: keep the radicand non-negative. $9 - x^2 \geq 0 \iff x^2 \leq 9 \iff -3 \leq x \leq 3$. Domain = $[-3, 3]$ (endpoints allowed: $\sqrt{0} = 0$ is fine).

Step 2 — range: track the radicand. As x runs over $[-3, 3]$, $x^2$ runs from 0 to 9, so $9 - x^2$ runs from 9 (at x = 0) down to 0 (at $x = \pm 3$).

Step 3 — take the square root. $\sqrt{\ }$ maps $[0, 9]$ onto $[0, 3]$: range = $[0, 3]$. Only the principal (non-negative) root counts.

Step 4 — geometric check. $y = \sqrt{9 - x^2}$ squares to $x^2 + y^2 = 9$ with $y \geq 0$ — the UPPER half of a circle of radius 3, confirming both domain and range.

Takeaway: even-root domains come from radicand $\geq 0$; the range follows by tracking the radicand's own range through the root.

Example 10. Find the domain and range of $f(x) = -|x|$.

Solution.

Step 1 — domain. $|x|$ is defined for every real number, and negation changes nothing: domain = $\mathbb{R}$.

Step 2 — range via the chain. $|x| \geq 0$ always, so $-|x| \leq 0$: outputs are confined to $(-\infty, 0]$.

Step 3 — every such value is attained. For any $y \leq 0$, take $x = -y \geq 0$: then $-|x| = -(-y) = y$. So range = $(-\infty, 0]$.

Takeaway: the graph is the V of $|x|$ flipped upside down — an inverted V peaking at the origin; negating a function reflects its range about 0.

Example 11. Find the range of: (i) $f(x) = 2 - 3x$, $x \in \mathbb{R}$, $x > 0$ (ii) $f(x) = x^2 + 2$ (iii) $f(x) = x$, x real.

Solution.

(i) Step 1 — chain the inequalities. $x > 0 \Rightarrow 3x > 0 \Rightarrow -3x < 0 \Rightarrow 2 - 3x < 2$. Every value below 2 is attained (solve $2 - 3x = y$ for any $y < 2$: $x = \frac{2-y}{3} > 0$ ✓). Range = $(-\infty, 2)$ — OPEN at 2 because x = 0 is excluded.

(ii) Step 2. $x^2 \geq 0$ with equality AT x = 0, so $x^2 + 2 \geq 2$ with the minimum 2 attained. Range = $[2, \infty)$ — CLOSED at 2.

(iii) Step 3. The identity function hits every real: range = $\mathbb{R}$.

Takeaway: track a chain of inequalities from the core piece, then decide open vs closed by asking whether the extreme is actually attained — (i) open, (ii) closed.

Example 12. The function f is defined by $f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}$. Describe its graph.

Solution.

Step 1 — plot the left piece. For x < 0, $f(x) = 1 - x$: at $x = -1, -2, -3$ the values are 2, 3, 4 — a line of slope $-1$ rising leftward from near (0, 1).

Step 2 — plot the right piece. For x > 0, $f(x) = x + 1$: values 2, 3, 4 at x = 1, 2, 3 — slope $+1$ rising rightward.

Step 3 — the joint. $f(0) = 1$ closes the gap: both pieces approach (0, 1), so the graph is one connected V with vertex (0, 1).

Step 4 — recognise the function. For $x < 0$: $1 - x = 1 + |x|$; for $x \geq 0$: $x + 1 = |x| + 1$. The whole formula is $f(x) = |x| + 1$ — the modulus V lifted one unit.

Takeaway: recognising a piecewise formula as a shifted modulus turns a graphing chore into a single sentence.

Example 13. The relation f is defined by $f(x) = \begin{cases} x^2, & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10 \end{cases}$ and the relation g by $g(x) = \begin{cases} x^2, & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10 \end{cases}$. Show that f is a function and g is not.

Solution.

Step 1 — locate the danger point. In a piecewise definition the only risk of \"two images for one input\" is where the pieces OVERLAP — here the shared boundary point.

Step 2 — test f at x = 3. First formula: $3^2 = 9$. Second: $3 \times 3 = 9$. The two claims AGREE, so x = 3 has the single image 9; every other x falls under exactly one formula. f is a function.

Step 3 — test g at x = 2. First formula: $2^2 = 4$. Second: $3 \times 2 = 6$. Two DIFFERENT images for the input 2. g is not a function.

Takeaway: for piecewise definitions, check agreement at every overlapping boundary — that single check decides the matter.

Example 14. Let f = {(1, 1), (2, 3), (0, $-1$), ($-1$, $-3$)} be a function from $\mathbb{Z}$ to $\mathbb{Z}$ defined by $f(x) = ax + b$. Determine a and b.

Solution.

Step 1 — pick the easiest data point. $(0, -1)$: $f(0) = a(0) + b = b = -1$.

Step 2 — use a second point. $(1, 1)$: $f(1) = a + b = 1$, and with $b = -1$: $a = 2$.

Step 3 — verify with the remaining points. $f(2) = 2(2) - 1 = 3$ ✓ matches (2, 3); $f(-1) = -2 - 1 = -3$ ✓ matches ($-1$, $-3$).

Step 4 — conclude. $f(x) = 2x - 1$, i.e. a = 2, b = $-1$.

Takeaway: two points determine a line — use x = 0 first if available (it isolates b instantly), and spend the leftover points on verification.

Example 15. Let f be the subset of $\mathbb{Z} \times \mathbb{Z}$ defined by f = {(ab, a + b) : a, b $\in \mathbb{Z}$}. Is f a function from $\mathbb{Z}$ to $\mathbb{Z}$?

Solution.

Step 1 — understand the pairs. Each pair has first entry the PRODUCT ab and second entry the SUM a + b. A function would require: same product ⟹ same sum, always.

Step 2 — hunt among factorisations. The number 4 factors two ways: $4 = 1 \times 4$ gives the pair (4, 5); $4 = 2 \times 2$ gives the pair (4, 4).

Step 3 — verdict. The input 4 has two different images, 5 and 4. Not a function.

Takeaway: to disprove \"function\", exhibit ONE input with two images — numbers with multiple factorisations are the natural hunting ground.

Example 16. Let A = {9, 10, 11, 12, 13} and $f : A \to \mathbb{N}$ be defined by f(n) = the highest prime factor of n. Find the range of f.

Solution.

Step 1 — factorise each element. $9 = 3^2$ → highest prime 3. $10 = 2 \times 5$ → 5. $11$ is prime → 11. $12 = 2^2 \times 3$ → 3. $13$ is prime → 13.

Step 2 — collect the images. 3, 5, 11, 3, 13 — as a set (repeats once): range = {3, 5, 11, 13}.

Takeaway: a function needs no algebraic formula — any well-defined rule qualifies; note 2 divides 10 and 12 yet never appears, because it is never the HIGHEST prime factor.