IB Mathematics AA – Topic 2: Functions

Key Definitions:

• Domain: The set of all possible input values (\(x\)-values) for which the function is defined

  Notation: \(D_f\) or \(\text{dom}(f)\)

• Range: The set of all possible output values (\(y\)-values) that the function can produce

  Notation: \(R_f\) or \(\text{range}(f)\)

Functions have restricted domains when:

• Division by zero: \(f(x) = \frac{1}{x}\) → \(x \neq 0\)
• Square roots (even roots): \(f(x) = \sqrt{x}\) → \(x \geq 0\)
• Logarithms: \(f(x) = \ln(x)\) → \(x > 0\)
• Tangent function: \(f(x) = \tan(x)\) → \(x \neq \frac{\pi}{2} + n\pi\)
• Explicitly stated: Function given with domain restriction

⚠ Common Pitfalls:

• Confusing domain and range: Domain is input (\(x\)), range is output (\(y\))
• Missing restrictions: Always check for division by zero, negative square roots, etc.
• Incorrect interval notation: Use \([a,b]\) for inclusive, \((a,b)\) for exclusive
• Forgetting asymptotes: Vertical asymptotes exclude values from domain

💡 Finding Range Tips:

• Sketch the graph to visualize output values
• Find minimum/maximum values (use calculus or GDC)
• Check end behavior: what happens as \(x \to \pm\infty\)?
• For composite functions: range of inner function becomes domain of outer

Example 1: Finding Domain and Range

Problem: Find the domain and range of:

(a) \(f(x) = \sqrt{4 - x^2}\)

(b) \(g(x) = \frac{x+1}{x-3}\)

Solution:

(a) \(f(x) = \sqrt{4 - x^2}\)

Domain: Square root requires non-negative argument

\(4 - x^2 \geq 0\)

\(x^2 \leq 4\)

\(-2 \leq x \leq 2\)

Domain: \(x \in [-2, 2]\)

Range: Consider the function's behavior

Maximum when \(x = 0\): \(f(0) = \sqrt{4} = 2\)

Minimum when \(x = \pm 2\): \(f(\pm 2) = \sqrt{0} = 0\)

Since \(\sqrt{\cdot} \geq 0\) and max is 2

Range: \(f(x) \in [0, 2]\)

(b) \(g(x) = \frac{x+1}{x-3}\)

Domain: Cannot divide by zero

\(x - 3 \neq 0\)

\(x \neq 3\)

Domain: \(x \in \mathbb{R} \setminus \{3\}\) or \(x \in (-\infty, 3) \cup (3, \infty)\)

Range: Let \(y = \frac{x+1}{x-3}\), solve for \(x\)

\(y(x-3) = x+1\)

\(yx - 3y = x + 1\)

\(yx - x = 3y + 1\)

\(x(y-1) = 3y + 1\)

\(x = \frac{3y+1}{y-1}\)

This is undefined when \(y = 1\)

Range: \(g(x) \in \mathbb{R} \setminus \{1\}\) or \(y \in (-\infty, 1) \cup (1, \infty)\)

Composite Function:

A composite function is created by applying one function to the output of another.

Order matters! \(f \circ g \neq g \circ f\) in general

Process: Start with \(x\), apply \(g\) first to get \(g(x)\), then apply \(f\) to that result

Finding Domain of \(f \circ g\):

1. \(x\) must be in the domain of \(g\)
2. \(g(x)\) must be in the domain of \(f\)

The domain of \(f \circ g\) is the set of \(x\) values where both conditions are satisfied

⚠ Common Mistakes:

• Wrong order: \(f(g(x))\) means apply \(g\) first, then \(f\)
• Forgetting to simplify: Expand and collect like terms
• Domain errors: Check restrictions from both functions
• Notation confusion: \(f \circ g \neq f \times g\)

Example 2: Composite Functions

Problem: Given \(f(x) = x^2 + 1\) and \(g(x) = 2x - 3\), find:

(a) \((f \circ g)(x)\)

(b) \((g \circ f)(x)\)

(c) \((f \circ g)(2)\)

Solution:

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

Start with \(f(x) = x^2 + 1\)

Replace \(x\) with \(g(x) = 2x - 3\):

\(f(g(x)) = (2x-3)^2 + 1\)

\(= 4x^2 - 12x + 9 + 1\)

\((f \circ g)(x) = 4x^2 - 12x + 10\)

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

Start with \(g(x) = 2x - 3\)

Replace \(x\) with \(f(x) = x^2 + 1\):

\(g(f(x)) = 2(x^2 + 1) - 3\)

\(= 2x^2 + 2 - 3\)

\((g \circ f)(x) = 2x^2 - 1\)

Note: \(f \circ g \neq g \circ f\) ✓

(c) \((f \circ g)(2)\)

Using result from (a): \((f \circ g)(x) = 4x^2 - 12x + 10\)

\((f \circ g)(2) = 4(2)^2 - 12(2) + 10\)

\(= 16 - 24 + 10\)

\((f \circ g)(2) = 2\)

Inverse Function \(f^{-1}\):

An inverse function reverses the effect of the original function

Important: \(f^{-1} \neq \frac{1}{f}\) (inverse is NOT reciprocal!)

One-to-One Requirement:

A function must be one-to-one (injective) for its inverse to exist

• Horizontal Line Test: Any horizontal line intersects the graph at most once
• Each output corresponds to exactly one input
• If not one-to-one, restrict the domain to make it one-to-one

Algorithm:

1. Write \(y = f(x)\)
2. Swap \(x\) and \(y\): write \(x = f(y)\)
3. Solve for \(y\) in terms of \(x\)
4. Replace \(y\) with \(f^{-1}(x)\)

Key Property:

• Domain of \(f\) = Range of \(f^{-1}\)
• Range of \(f\) = Domain of \(f^{-1}\)

Graphically: \(f^{-1}\) is the reflection of \(f\) across the line \(y = x\)

⚠ Common Errors:

• Confusing inverse with reciprocal: \(f^{-1}(x) \neq \frac{1}{f(x)}\)
• Forgetting to swap: Must exchange \(x\) and \(y\) before solving
• Not checking one-to-one: Inverse may not exist for all functions
• Domain/range swap: Remember they interchange

Example 3: Finding Inverse Functions

Problem: Find the inverse function and state its domain and range:

\(f(x) = \frac{2x + 3}{x - 1}\), \(x \neq 1\)

Solution:

Step 1: Write \(y = f(x)\)

\(y = \frac{2x + 3}{x - 1}\)

Step 2: Swap \(x\) and \(y\)

\(x = \frac{2y + 3}{y - 1}\)

Step 3: Solve for \(y\)

Multiply both sides by \((y - 1)\):

\(x(y - 1) = 2y + 3\)

\(xy - x = 2y + 3\)

Collect \(y\) terms:

\(xy - 2y = x + 3\)

\(y(x - 2) = x + 3\)

\(y = \frac{x + 3}{x - 2}\)

Step 4: Write inverse

\(f^{-1}(x) = \frac{x + 3}{x - 2}\), \(x \neq 2\)

Domain and Range:

For \(f\): Domain is \(x \in \mathbb{R} \setminus \{1\}\)

To find range of \(f\): as shown earlier, \(y \neq 2\)

So Range of \(f\) is \(y \in \mathbb{R} \setminus \{2\}\)

Domain of \(f^{-1}\): \(x \in \mathbb{R} \setminus \{2\}\) (range of \(f\))
Range of \(f^{-1}\): \(y \in \mathbb{R} \setminus \{1\}\) (domain of \(f\))

Note: Domain and range swap between function and inverse ✓

Definitions:

• Local Maximum: \(f(c)\) is a local max if \(f(c) \geq f(x)\) for all \(x\) near \(c\)
• Local Minimum: \(f(c)\) is a local min if \(f(c) \leq f(x)\) for all \(x\) near \(c\)
• Global (Absolute) Maximum: Largest value over entire domain
• Global (Absolute) Minimum: Smallest value over entire domain

Types of Intercepts:

• \(y\)-intercept: Point where graph crosses \(y\)-axis

  Set \(x = 0\) and solve for \(y\): \(y = f(0)\)

  Coordinates: \((0, f(0))\)

• \(x\)-intercept(s): Point(s) where graph crosses \(x\)-axis (also called roots or zeros)

  Set \(y = 0\) and solve for \(x\): \(f(x) = 0\)

  Coordinates: \((a, 0)\) where \(a\) is a solution

💡 Finding Extrema Tips:

• Use calculus: find where \(f'(x) = 0\) (critical points)
• Use GDC: graph and use maximum/minimum functions
• Check endpoints of domain for global extrema
• For parabolas: vertex is the extremum

Example 4: Intercepts and Extrema

Problem: For \(f(x) = x^2 - 4x + 3\), find:

(a) \(x\)-intercepts

(b) \(y\)-intercept

(c) Minimum value

Solution:

(a) \(x\)-intercepts: Set \(f(x) = 0\)

\(x^2 - 4x + 3 = 0\)

Factor: \((x - 1)(x - 3) = 0\)

\(x = 1\) or \(x = 3\)

\(x\)-intercepts: \((1, 0)\) and \((3, 0)\)

(b) \(y\)-intercept: Set \(x = 0\)

\(f(0) = 0^2 - 4(0) + 3 = 3\)

\(y\)-intercept: \((0, 3)\)

(c) Minimum value: (parabola opens upward, so vertex is minimum)

Complete the square or use vertex formula:

Vertex \(x\)-coordinate: \(x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2\)

Vertex \(y\)-coordinate: \(f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1\)

Minimum value: -1 at \(x = 2\)

Minimum point: \((2, -1)\)

To find where two functions intersect:

Set \(f(x) = g(x)\) and solve for \(x\)

Then find \(y\) by substituting back into either function

Steps for Accurate Sketching:

1. Identify domain and range (restrictions, asymptotes)
2. Find intercepts (\(x\) and \(y\) intercepts)
3. Find asymptotes (vertical: where undefined; horizontal: end behavior)
4. Locate extrema (maximum and minimum points)
5. Check symmetry (even function: \(f(-x) = f(x)\); odd: \(f(-x) = -f(x)\))
6. Plot key points and connect smoothly
7. Verify with GDC if allowed

💡 Sketching Tips:

• Always label axes and important points
• Show asymptotes as dashed lines
• Indicate clearly where function is undefined
• Use GDC to check your sketch
• Know standard function shapes (parabola, exponential, etc.)

Example 5: Complete Function Analysis (IB-Style)

Problem: For \(f(x) = \frac{x^2 - 4}{x - 1}\), \(x \neq 1\):

(a) Find the domain

(b) Find all intercepts

(c) Find any asymptotes

(d) Sketch the function

Solution:

(a) Domain:

Function undefined when \(x - 1 = 0\)

Domain: \(x \in \mathbb{R} \setminus \{1\}\)

(b) Intercepts:

\(x\)-intercepts: Set \(f(x) = 0\)

\(\frac{x^2 - 4}{x - 1} = 0\)

Numerator must be zero: \(x^2 - 4 = 0\)

\((x-2)(x+2) = 0\)

\(x = 2\) or \(x = -2\)

\(x\)-intercepts: \((-2, 0)\) and \((2, 0)\)

\(y\)-intercept: Set \(x = 0\)

\(f(0) = \frac{0^2 - 4}{0 - 1} = \frac{-4}{-1} = 4\)

\(y\)-intercept: \((0, 4)\)

(c) Asymptotes:

Vertical asymptote: Where denominator = 0

\(x = 1\)

Oblique asymptote: (degree of numerator > degree of denominator)

Use polynomial long division:

\(\frac{x^2 - 4}{x - 1} = x + 1 - \frac{3}{x-1}\)

As \(x \to \pm\infty\), \(\frac{3}{x-1} \to 0\)

Oblique asymptote: \(y = x + 1\)

(d) Sketch:

Key features to show:

• Vertical asymptote at \(x = 1\) (dashed line)
• Oblique asymptote \(y = x + 1\) (dashed line)
• Intercepts: \((-2, 0)\), \((0, 4)\), \((2, 0)\)
• Function approaches asymptotes at extremes

Note: Use GDC to verify sketch and identify behavior near asymptote

Common Question Types:

• "Find the domain/range": Check restrictions systematically
• "Find \(f \circ g\)": Apply inner function first, simplify
• "Find \(f^{-1}\)": Swap and solve, state domain/range
• "Sketch the function": Show intercepts, asymptotes, key points
• "Find where functions intersect": Set equal, solve

Calculator Usage:

• Use graph features to verify domain/range visually
• Find max/min using calculate menu
• Check intercepts with zero/root finder
• Verify inverse by checking \(f(f^{-1}(x)) = x\)