IB Mathematics AI – Topic 2

Notation:

Domain of f is written as: \(\text{Dom}(f)\) or \(D_f\)

Common Domain Restrictions:

1. Division by Zero:

For \(f(x) = \frac{g(x)}{h(x)}\), the denominator cannot be zero

Domain: all x where \(h(x) \neq 0\)

Example: \(f(x) = \frac{1}{x-3}\) has domain: \(x \neq 3\) or \((-\infty, 3) \cup (3, \infty)\)

2. Square Roots (Even Roots):

For \(f(x) = \sqrt{g(x)}\), the expression under the root must be non-negative

Domain: all x where \(g(x) \geq 0\)

Example: \(f(x) = \sqrt{x-2}\) has domain: \(x \geq 2\) or \([2, \infty)\)

3. Logarithms:

For \(f(x) = \log(g(x))\), the argument must be positive

Domain: all x where \(g(x) > 0\)

Example: \(f(x) = \ln(x+1)\) has domain: \(x > -1\) or \((-1, \infty)\)

Domain from Graph:

The domain is the projection of the graph onto the x-axis (horizontal extent)

Look for all x-values where the function exists

⚠️ Common Pitfalls & Tips:

• Use mnemonic "DEN": Denominator, Even roots, Natural logarithms
• Check ALL restrictions – a function may have multiple
• For composite functions, consider domain restrictions of both functions
• Use interval notation correctly: [ ] for inclusive, ( ) for exclusive
• Always verify answer with GDC graph

Notation:

Range of f is written as: \(\text{Range}(f)\) or \(R_f\)

Methods to Find Range:

1. Graphical Method:

Use GDC to graph the function

Range is the projection onto the y-axis (vertical extent)

Look for minimum and maximum y-values

2. Algebraic Method:

Rearrange equation to solve for x in terms of y

Find restrictions on y that make x real and in the domain

3. Analysis Method:

For quadratics: Find vertex using \(x = -\frac{b}{2a}\)

Consider function behavior and asymptotes

Use calculus to find maximum/minimum points

Common Function Ranges:

• \(f(x) = x^2\): Range is \([0, \infty)\)
• \(f(x) = e^x\): Range is \((0, \infty)\)
• \(f(x) = \sin(x)\): Range is \([-1, 1]\)
• \(f(x) = \ln(x)\): Range is \((-\infty, \infty)\)
• \(f(x) = \frac{1}{x}\): Range is \((-\infty, 0) \cup (0, \infty)\)

⚠️ Common Pitfalls & Tips:

• Range is often harder to find than domain – use GDC!
• Don't assume range is all real numbers
• For restricted domains, range may also be restricted
• Check endpoints of domain for range boundaries

Solution:

Finding Domain:

For square root to be defined, expression inside must be ≥ 0:

\[ 4 - x^2 \geq 0 \]

\[ 4 \geq x^2 \]

\[ -2 \leq x \leq 2 \]

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

Finding Range:

The function represents the upper semicircle of radius 2

Minimum value: \(f(\pm 2) = \sqrt{4-4} = 0\)

Maximum value: \(f(0) = \sqrt{4-0} = 2\)

Range: \([0, 2]\)

Verification using graph:

Graph shows semicircle from x = -2 to x = 2

Height varies from y = 0 to y = 2

Notation:

\[ (f \circ g)(x) = f(g(x)) \]

Read as: "f composed with g" or "f of g of x"

Process: First apply g to x, then apply f to the result

Key Properties:

• NOT commutative: \(f \circ g \neq g \circ f\) (usually)
• Order matters: Apply functions from right to left
• Associative: \(f \circ (g \circ h) = (f \circ g) \circ h\)

Steps to Evaluate \((f \circ g)(x)\):

1. Start with the inner function: find \(g(x)\)
2. Substitute result into outer function: find \(f(g(x))\)
3. Simplify the expression

Domain of Composite Functions:

For \((f \circ g)(x) = f(g(x))\):

• x must be in the domain of g
• g(x) must be in the domain of f
• Domain is the intersection of these conditions

Example:

If \(f(x) = x^2\) and \(g(x) = x + 3\):

\[ (f \circ g)(x) = f(g(x)) = f(x+3) = (x+3)^2 = x^2 + 6x + 9 \]

\[ (g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 3 \]

Note: \((f \circ g)(x) \neq (g \circ f)(x)\)

⚠️ Common Pitfalls & Tips:

• CRITICAL: Work from right to left, inside to outside
• Don't confuse \(f \circ g\) with \(f \cdot g\) (multiplication)
• Check domain restrictions carefully for composite functions
• Show all steps – don't skip the intermediate function

Solution:

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

Step 1: Find \(g(x) = x^2 + 3\)

Step 2: Substitute into f:

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

\[ = 2x^2 + 6 - 1 = 2x^2 + 5 \]

Answer: \((f \circ g)(x) = 2x^2 + 5\)

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

Step 1: Find \(f(x) = 2x - 1\)

Step 2: Substitute into g:

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

\[ = 4x^2 - 4x + 1 + 3 = 4x^2 - 4x + 4 \]

Answer: \((g \circ f)(x) = 4x^2 - 4x + 4\)

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

Using result from (a):

\[ (f \circ g)(2) = 2(2)^2 + 5 = 2(4) + 5 = 13 \]

Answer: 13

Key Relationship:

\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \]

Or equivalently: \((f \circ f^{-1})(x) = x\) and \((f^{-1} \circ f)(x) = x\)

Important Note:

\(f^{-1}(x)\) is NOT the same as \(\frac{1}{f(x)}\)

\(f^{-1}\) means inverse function, not reciprocal

When Does an Inverse Exist?

A function has an inverse if and only if it is one-to-one (injective)

• Horizontal Line Test: If any horizontal line intersects the graph more than once, the function does NOT have an inverse
• Each y-value corresponds to exactly one x-value

Steps to Find Inverse Function:

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

Properties of Inverse Functions:

• Domain and Range swap: Dom(\(f^{-1}\)) = Range(f), Range(\(f^{-1}\)) = Dom(f)
• Graphs are reflections: Graph of \(f^{-1}\) is reflection of f over line y = x
• Symmetry: If point (a, b) is on f, then (b, a) is on \(f^{-1}\)

⚠️ Common Pitfalls & Tips:

• \(f^{-1}(x) \neq \frac{1}{f(x)}\) – not the reciprocal!
• Always verify: check that \(f(f^{-1}(x)) = x\)
• Not all functions have inverses (must be one-to-one)
• Domain restrictions may be needed for inverse to exist
• Use GDC to verify inverse graphically (reflection test)

Solution:

Step 1: Replace f(x) with y

\[ y = \frac{2x + 3}{x - 1} \]

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 on one side:

\[ xy - 2y = x + 3 \]

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

\[ y = \frac{x + 3}{x - 2} \]

Step 4: Write inverse function

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

Verification:

Check \(f(f^{-1}(x)) = x\):

\[ f(f^{-1}(x)) = f\left(\frac{x+3}{x-2}\right) = \frac{2\left(\frac{x+3}{x-2}\right) + 3}{\frac{x+3}{x-2} - 1} \]

\[ = \frac{\frac{2x+6+3x-6}{x-2}}{\frac{x+3-x+2}{x-2}} = \frac{5x}{5} = x \checkmark \]

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

Domain & Range

• Domain: All possible x-values
• Range: All possible y-values
• Check: denominators, roots, logs
• Use GDC to verify

Composite Functions

• \((f \circ g)(x) = f(g(x))\)
• Work inside-out (right to left)
• Usually NOT commutative
• Check domain restrictions

Inverse Functions

• \(f(f^{-1}(x)) = x\)
• Swap x and y, then solve
• Must be one-to-one
• Domain/Range swap

Key Tests

• Vertical Line Test: Is it a function?
• Horizontal Line Test: Does inverse exist?
• Both needed for full analysis