Functions and Graphs — Reading, Sketching, Transforming

What Is a Function, Really?

A function is a rule that assigns to each input exactly one output. The word “exactly one” is the critical part — if any input maps to two different outputs, it’s not a function.

Think of it like a machine: you put something in, exactly one thing comes out. Every vending machine is a function — press button A1, get one specific snack.

Formally, a function $f: A \to B$ maps every element of set $A$ (domain) to exactly one element of set $B$ (co-domain). The set of actual output values is the range (which may be smaller than the co-domain).

Domain, Range, and Co-domain

Domain: Set of all valid inputs. When not specified, assume all real numbers for which the expression is defined.

Co-domain: The set the function maps into (often $\mathbb{R}$ by default).

Range: The set of actual outputs — the image of the domain under $f$.

Example: $f(x) = \frac{1}{\sqrt{x-2}}$

• Denominator cannot be 0: $\sqrt{x-2} \neq 0 \Rightarrow x \neq 2$
• Under square root must be non-negative: $x - 2 \geq 0 \Rightarrow x \geq 2$
• Combined: $x > 2$, so Domain $= (2, \infty)$

Types of Functions

Injective (One-One) Functions

$f$ is injective if different inputs give different outputs: $f(a) = f(b) \Rightarrow a = b$.

Horizontal line test: If any horizontal line crosses the graph more than once, the function is not injective.

Surjective (Onto) Functions

$f: A \to B$ is surjective if every element of $B$ is actually achieved: Range = Co-domain.

Bijective Functions

Both injective and surjective. Only bijective functions have well-defined inverses.

Standard Functions and Their Graphs

1. Constant Function: $f(x) = c$

Horizontal line at height $c$. Domain: $\mathbb{R}$, Range: $\{c\}$. Not injective (all inputs give same output), technically surjective only if co-domain is $\{c\}$.

2. Identity Function: $f(x) = x$

Line through origin with slope 1. Domain and Range both $\mathbb{R}$. Bijective.

3. Linear Function: $f(x) = mx + c$

Straight line. Bijective for $m \neq 0$. Slope $m$ tells us steepness; $c$ is the y-intercept.

4. Quadratic Function: $f(x) = ax^2 + bx + c$

Parabola. If $a > 0$, opens upward (minimum at vertex). If $a < 0$, opens downward (maximum at vertex).

Vertex: $x = -\frac{b}{2a}$, $y = c - \frac{b^2}{4a}$

Not injective over full $\mathbb{R}$ (symmetric about vertex), but injective if restricted to one half.

5. Modulus Function: $f(x) = |x|$

V-shape with vertex at origin. Domain: $\mathbb{R}$, Range: $[0, \infty)$. Not injective.

6. Greatest Integer Function (Floor): $f(x) = \lfloor x \rfloor$

Also written $[x]$. Returns the greatest integer $\leq x$. Graph is a staircase with closed left endpoint and open right endpoint on each step.

$[2.7] = 2$, $[-1.3] = -2$, $[3] = 3$

7. Signum Function

Range: $\{-1, 0, 1\}$.

8. Exponential Function: $f(x) = a^x$, $a > 0, a \neq 1$

Domain: $\mathbb{R}$, Range: $(0, \infty)$. Always passes through $(0, 1)$.

• If $a > 1$: increasing (standard growth curve)
• If $0 < a < 1$: decreasing (decay curve)

9. Logarithmic Function: $f(x) = \log_a x$

Domain: $(0, \infty)$, Range: $\mathbb{R}$. Inverse of exponential. Passes through $(1, 0)$.

Graph Transformations

Understanding transformations lets you sketch complex functions without plotting dozens of points.

Starting from a base graph $y = f(x)$:

Worked Example: Sketch $y = |x^2 - 4|$

Step 1: Sketch $y = x^2 - 4$ (parabola, vertex at $(0,-4)$, crosses x-axis at $\pm 2$).

Step 2: Any part of the curve below the x-axis gets reflected above it.

Step 3: The portion between $x = -2$ and $x = 2$ (where $x^2 - 4 < 0$) flips upward, creating a “W” shape.

Composition and Inverse

Composition $f \circ g$

$(f \circ g)(x) = f(g(x))$

We apply $g$ first, then $f$. The domain of $f \circ g$ is the set of $x$ where $g(x)$ is in the domain of $f$.

Inverse Function $f^{-1}$

If $f$ is bijective, then $f^{-1}$ exists. The graph of $f^{-1}$ is the mirror image of $f$ about the line $y = x$.

Finding $f^{-1}$: Replace $f(x)$ with $y$, swap $x$ and $y$, solve for $y$.

Example: $f(x) = 3x - 5$. Let $y = 3x - 5$. Swap: $x = 3y - 5$. Solve: $y = \frac{x+5}{3}$. So $f^{-1}(x) = \frac{x+5}{3}$.

Verification: $f(f^{-1}(x)) = 3 \cdot \frac{x+5}{3} - 5 = x$. Correct.

Even and Odd Functions

Even function: $f(-x) = f(x)$ for all $x$ in the domain. Graph is symmetric about the y-axis. Examples: $x^2, |x|, \cos x$.

Odd function: $f(-x) = -f(x)$ for all $x$ in the domain. Graph has rotational symmetry about the origin. Examples: $x, x^3, \sin x$.

Neither: Most functions. Example: $f(x) = x^2 + x$.

Solved Examples

Example 1 — CBSE Level

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

We need $9 - x^2 \geq 0 \Rightarrow x^2 \leq 9 \Rightarrow -3 \leq x \leq 3$.

Domain $= [-3, 3]$

Example 2 — JEE Main Level

If $f(x) = \frac{x}{1+x^2}$, check if it’s even, odd, or neither. Also find its range.

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

For range: Let $y = \frac{x}{1+x^2}$. Then $yx^2 - x + y = 0$. For real $x$: discriminant $\geq 0$: $1 - 4y^2 \geq 0 \Rightarrow y^2 \leq \frac{1}{4} \Rightarrow -\frac{1}{2} \leq y \leq \frac{1}{2}$.

Range $= \left[-\frac{1}{2}, \frac{1}{2}\right]$

Example 3 — JEE Advanced Level

$f: \mathbb{R} \to \mathbb{R}$ is given by $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ and $f(1) = 2$. Find $f(n)$ for positive integer $n$, and show that $\sum_{k=1}^{n} f(k)$ forms a geometric series.

Setting $x = y = 0$: $f(0) = f(0)^2 \Rightarrow f(0) = 0$ or $1$. Since $f(1) = 2 \neq 0$, $f(0) = 1$.

Setting $y = 1$: $f(x+1) = f(x) \cdot f(1) = 2f(x)$.

So $f(2) = 2f(1) = 4$, $f(3) = 2f(2) = 8$, …, $f(n) = 2^n$.

Sum $= 2 + 4 + 8 + \cdots + 2^n = 2(2^n - 1)/(2-1) = 2^{n+1} - 2$.

Common Mistakes to Avoid

Practice Questions

Q1. Find the domain and range of $f(x) = \frac{x-1}{x+2}$.

Q2. If $f(x) = x^2 + 1$ and $g(x) = 2x - 1$, find $(f \circ g)(2)$.

Q3. Is $f(x) = x|x|$ even, odd, or neither?

Q4. Find the inverse of $f(x) = \frac{2x+1}{x-3}$.

Q5. Evaluate $[3.7] + [-1.2] + [0] + [-2.9]$.

Q6. Describe the transformation from $y = x^2$ to $y = -(x-3)^2 + 4$.

FAQs

Q: What is the difference between “one-one” and “many-one” functions?

One-one (injective): distinct inputs always give distinct outputs. Many-one: at least two different inputs give the same output. $f(x) = x^2$ is many-one since $f(2) = f(-2) = 4$.

Q: Can a function be both even and odd?

Yes — only $f(x) = 0$ (the zero function) satisfies both $f(-x) = f(x)$ and $f(-x) = -f(x)$ simultaneously.

Q: In JEE, how often does greatest integer function appear?

Almost every year. Common question types: finding range of $[f(x)]$, evaluating limits involving $[x]$, and solving equations like $[x^2 - 4] = 0$.

Q: How do I prove a function is bijective in CBSE?

Standard two-step proof: (1) Assume $f(a) = f(b)$ and show $a = b$ (injectivity). (2) For any $y$ in co-domain, find $x$ in domain with $f(x) = y$ (surjectivity). Always show both steps explicitly.

Q: What is the significance of the vertical line test?

If any vertical line crosses the graph more than once, the curve does not represent a function — some x-value maps to multiple y-values. This visually checks the “exactly one output per input” rule.