Functions – Domain, Range, Types & Real‑World Applications

1. What is a Function?

A function $f$ from a set $A$ (the domain) to a set $B$ (the codomain) is a rule that assigns to every $x\in A$ a unique element $y\in B$.

• Notation: $y=f(x)$ or $f:A\to B$.
• Image (range): $\mathrm{Im}(f)=\{f(x)\mid x\in A\}$.

Visual: Function as a Mapping Diagram

flowchart LR
    subgraph "Domain A"
        A1["-2"] -->|"f"| B1["5"]
        A2["0"] -->|"f"| B2["1"]
        A3["2"] -->|"f"| B3["9"]
    end
    subgraph "Codomain B"
        B1
        B2
        B3
    end

2. Determining the Domain

The domain consists of all real numbers for which the expression defining $f$ is legal:

Example 1 – Domain of a Rational‑Root Function

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

Steps

1. Denominator cannot be zero: $3-x\ne 0\Rightarrow x\ne 3$.
2. No other restrictions.

$$\boxed{\text{Domain }{D}_f=\mathbb{R}\setminus \{3\}}.$$

3. Determining the Range

Two common strategies:

1. Solve for $x$ in terms of $y$ and apply domain restrictions to the new expression.
2. Complete the square (for quadratic radicands) or use calculus (derivative) to locate extrema.

Worked Example 2 – Range of a Quadratic‑Root Function

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

Step 1 – Domain (radicand $\ge 0$)

$$2-x-x^2\ge 0⟺-(x^2+x-2)\ge 0⟺x^2+x-2\le 0.$$

Factor: $x^2+x-2=(x+2)(x-1)$.
Sign chart gives $-2\le x\le 1$.

$$D_f=[-2,1].$$

Step 2 – Range

Let $y=\sqrt{2-x-x^2}$ with $y\ge 0$. Square both sides:

$$y^2=2-x-x^2⟹x^2+x+(y^2-2)=0.$$

Treat as quadratic in $x$. Real solutions exist iff discriminant $\Delta \ge 0$:

$$\Delta =1^2-4\cdot 1\cdot (y^2-2)=1-4y^2+8=9-4y^2\ge 0$$ $$⟹4y^2\le 9⟹y^2\le \frac{9}{4}⟹0\le y\le \frac{3}{2}.$$

Thus

$$\boxed{\text{Range }{R}_f=[0,\frac{3}{2}]}.$$

Visual: Graph of $f(x)=\sqrt{2-x-x^2}$

Visual: Number‑line showing domain

4. Types of Functions

Visual: Venn diagram of function types

5. Proving Bijectivity – A Sample Proof

Statement: Show that $f:\mathbb{R}\to \mathbb{R},f(x)=3x+5$ is bijective.

Injective proof:
Assume $f(x_1)=f(x_2)$.
$3x_1+5=3x_2+5\Rightarrow 3x_1=3x_2\Rightarrow x_1=x_2$. Hence injective.

Surjective proof:
Take any $y\in \mathbb{R}$. Solve $y=3x+5$ for $x$:
$x=\frac{y-5}{3}$, which is a real number for any real $y$. Hence every $y$ has a pre‑image; surjective.

Since both hold, $f$ is bijective and possesses an inverse
$$f^{-1}(y)=\frac{y-5}{3}.$$

6. Composite and Inverse Functions

Given $f:A\to B$ and $g:B\to C$, the composition $g\circ f:A\to C$ is defined by $(g\circ f)(x)=g(f(x))$.

If $f$ is bijective, the inverse $f^{-1}:B\to A$ satisfies
$f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$.

Worked Example 3 – Composite & Inverse

Let $f(x)=2x+1$ and $g(x)=x^2-2$.

• Find $(g\circ f)(x)$.
  $$(g\circ f)(x)=g(f(x))=(2x+1{)}^2-2=4x^2+4x+1-2=4x^2+4x-1.$$

• Determine if $f$ has an inverse and write it.
  $f$ is linear with slope 2 > 0 ⇒ injective and surjective on $\mathbb{R}$.
  $f^{-1}(y)=\frac{y-1}{2}$.

7. Graphical Interpretation of Domain & Range

A function’s graph $y=f(x)$ visually displays the admissible $x$-values (where the curve exists) and the corresponding $y$-values (vertical spread).

Visual: Graph of a rational‑root function

The vertical line $x=3$ is a domain restriction (hole). The curve approaches it but never touches, confirming the domain $\mathbb{R}\setminus \{3\}$. The range is all real numbers except the value that would make the numerator zero when denominator is zero – here the range is $\mathbb{R}\setminus \{2\}$ (students can verify by solving $y=\frac{2x+1}{3-x}$ for $x$ and checking the forbidden $y$).

8. Comparison Table – Quick Reference

| Property | Test (Algebraic) | Graphical clue |
|----------|------------------|----------------|
| Injective | \(f(x_1)=f(x_2)\Rightarrow x_1=x_2\) | Passes Horizontal Line Test |
| Surjective | Solve \(y=f(x)\) for arbitrary \(y\) | Curve covers entire codomain vertically |
| Bijective | Both of the above | Passes Horizontal Test **and** covers codomain |
| Even | \(f(-x)=f(x)\) | Symmetric about y‑axis |
| Odd | \(f(-x)=-f(x)\) | Rotational symmetry 180° about origin |
| Periodic | Find smallest \(p>0\) with \(f(x+p)=f(x)\) | Repeating pattern on graph |

9. Applications in Everyday Technology

9.1 eSewa Payment Mapping

Product: eSewa mobile wallet.
Function idea used: Injective mapping from transaction ID → unique payment record. Each ID maps to exactly one record, preventing duplicate payments.

9.2 Daraz Order Queue

Product: Daraz e‑commerce order processing.
Function idea used: Surjective mapping from order status (Pending, Packed, Shipped, Delivered) → customer notifications. Every possible status appears in at least one notification, ensuring customers are always informed.

9.3 Ncell Signal Strength

Product: Ncell mobile network.
Function idea used: Bijective mapping between cell‑tower coordinates and signal‑strength values for a given user location (within a small cell). The inverse function lets the app compute the nearest tower from a measured signal strength.

10. In the real world

• eSewa uses an injective function $f(\text{TxnID})=\text{PaymentRecord}$. If two different transaction IDs produced the same record, fraud could occur. The system guarantees $f$ is one‑to‑one by attaching a cryptographic hash to each ID.
• Daraz maintains a surjective function $g(\text{OrderStatus})=\text{NotificationMessage}$. Every status (e.g., “Shipped”) has at least one corresponding message (“Your order has been shipped”). The platform checks that no status is left without a notification, ensuring full coverage.
• Ncell implements a bijective mapping $h(\text{TowerCoord})=\text{SignalLevel}$ within a micro‑cell. Because the mapping is bijective, the app can invert it: given a measured signal level, it can uniquely locate the serving tower, which is essential for hand‑over decisions.

Worked Real‑World Example

A Nepali bank offers a personal loan where the interest payable $I$ depends on the loan amount $L$ via
$$I(L)=\frac{5}{100}L+2000.$$
This is a linear function $f(L)=0.05L+2000$.

• Domain: All permissible loan amounts, e.g., $L\in [50,000,5,000,000]$ Rs.
• Range: Corresponding interest values, computed by plugging domain endpoints:
  $$I_{\min }=0.05(50,000)+2000=4250,I_{\max }=0.05(5,000,000)+2000=252,000.$$
• The function is bijective on this interval, so the bank can invert it to find the loan amount from a desired interest payment:
  $$L=\frac{I-2000}{0.05}.$$

This mirrors the textbook bijectivity proof for $f(x)=3x+5$.

11. Common Pitfalls

12. Practice Problems (with brief solutions)

1. Domain & Range: $f(x)=\sqrt{9-x^2}$.

   • Domain: $[-3,3]$.
   • Range: $[0,3]$.

2. Injectivity Test: $f(x)=x^3-x$. Show it is not injective by finding two distinct $x$ with same output (e.g., $x=0$ and $x=1$ give $0$ and $0$ respectively).

3. Surjectivity on $\mathbb{R}$: $f(x)=e^x$. Not surjective because negative numbers are never attained.

4. Inverse Function: Find $f^{-1}(x)$ for $f(x)=\frac{1}{x-2}$.

   • Solve $y=\frac{1}{x-2}\Rightarrow x=2+\frac{1}{y}$.
   • Hence $f^{-1}(y)=2+\frac{1}{y}$ with domain $y\ne 0$.

13. Exam tip

• Read the question carefully: “Find domain” → list all algebraic restrictions; “Find range” → either solve for $x$ in terms of $y$ or complete the square.
• Show every step: write the inequality, factor, draw a quick sign chart, then state the interval. Marks are awarded for the reasoning, not just the final interval.
• For bijectivity, provide both an injective proof (usually by assuming $f(x_1)=f(x_2)$) and a surjective proof (solve $y=f(x)$ for arbitrary $y$).
• Graph sketches earn extra points: a neat curve with clearly marked domain endpoints and range extremes demonstrates understanding.
• Watch out for hidden restrictions in composite functions—ensure the inner function’s range lies inside the outer function’s domain.