MYP Inquiry Framework
Key Concept Relationships
Related Concepts Representation, Transformation, Space
Global Context Personal and cultural expression (Symmetry, computer graphics & signal processing)
Statement of Inquiry Mathematical transformations alter the spatial representation of functions while maintaining fundamental algebraic relationships.
6.1

Definition of a Function, Domain & Range

  • Relation vs Function: A relation is any set of ordered pairs $(x, y)$. A function $f$ is a special relation where each input $x$ in the domain maps to exactly one output $y$ in the range.
  • Vertical Line Test (VLT): A graph represents a function if and only if no vertical line crosses the graph more than once.
  • Domain Restrictions: The set of all allowable real inputs:
    • Denominators cannot equal zero (e.g., $f(x) = \frac{1}{x-4} \implies x \ne 4$).
    • Even roots cannot take negative radicands (e.g., $f(x) = \sqrt{2x - 6} \implies 2x - 6 \ge 0 \implies x \ge 3$).
  • Range: The set of all possible resulting output values $f(x)$ as $x$ varies over the entire domain.
Function Machine f(x) x f(x) Inverse: Reflection in y = x y = x f(x) f⁻¹(x)
6.2

Composite Functions: $(f \circ g)(x) = f(g(x))$

  • Composition Concept: Applying one function to the result of another. The output of the inner function $g(x)$ becomes the input for the outer function $f$: $$(f \circ g)(x) = f(g(x))$$
  • Non-Commutative Property: In general, the order of composition matters: $$f(g(x)) \ne g(f(x))$$
  • Domain of Composite Function: For $f(g(x))$ to exist, $x$ must be in the domain of $g$, and $g(x)$ must lie within the allowable domain of $f$.
6.3

Inverse Functions: $f^{-1}(x)$

  • Condition for Existence: An inverse function $f^{-1}(x)$ exists if and only if $f$ is a one-to-one (bijective) function, meaning it passes the Horizontal Line Test (HLT). If not one-to-one, the domain must be restricted (e.g., $f(x) = x^2$ restricted to $x \ge 0$).
  • Algebraic Algorithm for Finding $f^{-1}(x)$:
    1. Write the function as $y = f(x)$.
    2. Swap the variables $x$ and $y$: $x = f(y)$.
    3. Solve this equation algebraically for $y$ in terms of $x$.
    4. Replace $y$ with $f^{-1}(x)$.
  • Fundamental Identity: $$(f \circ f^{-1})(x) = f(f^{-1}(x)) = x \quad \text{and} \quad (f^{-1} \circ f)(x) = x$$
  • Geometric Property: The graph of $f^{-1}(x)$ is the reflection of $y = f(x)$ across the identity line $\mathbf{y = x}$. $$\text{Domain}(f^{-1}) = \text{Range}(f), \qquad \text{Range}(f^{-1}) = \text{Domain}(f)$$
6.4

Geometric Graph Transformations

  • Vertical Translation: $y = f(x) + k$ Translates the graph vertically by vector $\begin{pmatrix} 0 \\ k \end{pmatrix}$. Shifts up by $k$ if $k > 0$, down if $k < 0$.
  • Horizontal Translation: $y = f(x - h)$ Translates the graph horizontally by vector $\begin{pmatrix} h \\ 0 \end{pmatrix}$. Shifts right by $h$ if $(x - h)$, and left if $(x + h)$.
  • Reflections:
    • $y = -f(x)$: Reflection in the $x$-axis (vertical inversion, $(x, y) \to (x, -y)$).
    • $y = f(-x)$: Reflection in the $y$-axis (horizontal inversion, $(x, y) \to (-x, y)$).
  • Stretches (Dilations):
    • $y = a \cdot f(x)$: Vertical stretch by scale factor $|a|$ parallel to $y$-axis.
    • $y = f(bx)$: Horizontal stretch by scale factor $\frac{1}{|b|}$ parallel to $x$-axis.

Worked Examples (Criterion A & C)

Example 1 Composite Functions & Domain Restrictions

Problem: Given $f(x) = \dfrac{2}{x - 3}$ ($x \ne 3$) and $g(x) = x^2 + 1$.
(a) Find $(f \circ g)(x)$ and state its domain.
(b) Evaluate $(g \circ f)(5)$.

Step 1: Determine $(f \circ g)(x) = f(g(x))$ $$f(g(x)) = f(x^2 + 1) = \frac{2}{(x^2 + 1) - 3} = \mathbf{\frac{2}{x^2 - 2}}$$
Step 2: Determine domain of $(f \circ g)(x)$

The denominator cannot be zero: $x^2 - 2 \ne 0 \implies x^2 \ne 2 \implies x \ne \pm \sqrt{2}$.

Domain: $\mathbf{\{x \in \mathbb{R} \mid x \ne \pm\sqrt{2}\}}$.

Step 3: Evaluate $(g \circ f)(5) = g(f(5))$

First find $f(5)$: $f(5) = \frac{2}{5 - 3} = \frac{2}{2} = 1$.

Then evaluate $g(1)$: $g(1) = 1^2 + 1 = \mathbf{2}$.

Criterion A: Correct substitution and precise algebraic domain restrictions
Example 2 Finding Rational Inverse Function

Problem: Find the inverse function of $f(x) = \dfrac{3x + 1}{x - 2}$ ($x \ne 2$).

Step 1: Set $y = f(x)$ and swap variables $$y = \frac{3x + 1}{x - 2} \implies x = \frac{3y + 1}{y - 2}$$
Step 2: Clear denominator and expand $$x(y - 2) = 3y + 1 \implies xy - 2x = 3y + 1$$
Step 3: Collect terms in $y$ on one side and factorise $$xy - 3y = 2x + 1 \implies y(x - 3) = 2x + 1$$
Step 4: Solve for $y$ to obtain $f^{-1}(x)$ $$\mathbf{f^{-1}(x) = \frac{2x + 1}{x - 3}} \quad (x \ne 3)$$ Criterion C: Clear line-by-line deduction showing literal inversion

MYP 4–5 Practice Exercises: Functions & Transformations

Tier 1 (Core skills), Tier 2 (Intermediate Reasoning), and Tier 3 (Extended & Criteria B/C Investigations).

Question 1 Tier 1: Standard 3 Marks

Given $f(x) = 4 - 3x^2$, calculate:
(a) $f(-2)$
(b) $f(0)$

View Full Method Markscheme
[M1] Part (a): Substitute $x = -2$: $f(-2) = 4 - 3(-2)^2 = 4 - 3(4)$.
[A1] $f(-2) = 4 - 12 = \mathbf{-8}$.
[A1] Part (b): Substitute $x = 0$: $f(0) = 4 - 3(0) = \mathbf{4}$.
Question 2 Tier 1: Standard 3 Marks

State the natural domain of the real-valued function $g(x) = \sqrt{5x - 15}$.

View Full Method Markscheme
[M1] For a real square root, the expression under the radical must be non-negative: $5x - 15 \ge 0$.
[M1] Solve inequality: $5x \ge 15 \implies x \ge 3$.
[A1] Domain: $\mathbf{\{x \in \mathbb{R} \mid x \ge 3\}}$ or in interval notation $\mathbf{[3, \infty)}$.
Question 3 Tier 1: Standard 3 Marks

The graph of $y = f(x)$ is translated $4\text{ units to the right}$ and $3\text{ units downwards}$. Write down the equation of the transformed graph in terms of $f$.

View Full Method Markscheme
[M1] Horizontal shift right by $4$: replace $x$ with $(x - 4)$.
[M1] Vertical shift down by $3$: subtract $3$ from the overall expression.
[A1] Final equation: $\mathbf{y = f(x - 4) - 3}$.
Question 4 Tier 2: Intermediate 4 Marks

Find the inverse function $f^{-1}(x)$ for $f(x) = \dfrac{5x - 7}{2}$.

View Full Method Markscheme
[M1] Set $y = \frac{5x - 7}{2}$ and interchange $x$ and $y$: $x = \frac{5y - 7}{2}$.
[M1] Multiply by $2$: $2x = 5y - 7$.
[M1] Add $7$: $5y = 2x + 7$.
[A1] Isolate $y$: $\mathbf{f^{-1}(x) = \dfrac{2x + 7}{5}}$.
Question 5 Tier 2: Intermediate 4 Marks

Let $f(x) = 2x + 3$ and $g(x) = x^2 - 4$.
(a) Find an expression for $(f \circ g)(x)$.
(b) Solve the equation $(f \circ g)(x) = 11$.

View Full Method Markscheme
[M1] Part (a): $(f \circ g)(x) = f(g(x)) = 2(x^2 - 4) + 3 = 2x^2 - 8 + 3 = \mathbf{2x^2 - 5}$.
[M1] Part (b): Set $2x^2 - 5 = 11$.
[M1] Add $5$: $2x^2 = 16 \implies x^2 = 8$.
[A1] Solve: $\mathbf{x = \pm \sqrt{8} = \pm 2\sqrt{2}}$.
Question 6 Tier 2: Intermediate 4 Marks

The vertex of a parabola $y = f(x)$ is at $(-3, 5)$. Find the new vertex coordinates after applying the transformation $y = -2f(x + 1) + 4$.

View Full Method Markscheme
[M1] Horizontal shift $f(x + 1)$: moves left by $1$, so new $x$-coordinate is $-3 - 1 = \mathbf{-4}$.
[M1] Vertical reflection and stretch by $-2$: multiply $y$-coordinate by $-2$, giving $5 \times (-2) = -10$.
[M1] Vertical shift $+4$: add $4$ to $y$, giving $-10 + 4 = \mathbf{-6}$.
[A1] New vertex coordinates: $\mathbf{(-4, -6)}$.
Question 7 Tier 2: Intermediate 4 Marks

Show that the function $f(x) = \dfrac{4x - 1}{x - 4}$ ($x \ne 4$) is self-inverse, meaning $f^{-1}(x) = f(x)$.

View Full Method Markscheme
[M1] Set $x = \frac{4y - 1}{y - 4}$. Clear fractions: $x(y - 4) = 4y - 1$.
[M1] Expand: $xy - 4x = 4y - 1$.
[M1] Group $y$: $xy - 4y = 4x - 1 \implies y(x - 4) = 4x - 1$.
[A1] Isolate $y$: $y = \frac{4x - 1}{x - 4} = f(x)$. Thus $\mathbf{f^{-1}(x) = f(x)}$. [Proved]
Question 8 Tier 3: Extended 5 Marks

Consider $f(x) = x^2 - 6x + 13$.
(a) Explain why $f$ does not have an inverse for all $x \in \mathbb{R}$.
(b) State the domain restriction $x \ge k$ with the smallest possible integer $k$ for which an inverse exists.
(c) Find the equation for $f^{-1}(x)$ for this restricted domain, including its domain.

View Full Method Markscheme
[M1] Part (a): $f(x)$ is a parabola; it fails the Horizontal Line Test because multiple $x$-values yield the same $y$-value (many-to-one relation).
[M1] Part (b): Complete the square: $f(x) = (x - 3)^2 + 4$. The vertex is at $x = 3$. Therefore, restricting the domain to the right branch $x \ge 3$ makes it one-to-one. So $\mathbf{k = 3}$.
[M1] Part (c): For $x \ge 3$, set $x = (y - 3)^2 + 4 \implies (y - 3)^2 = x - 4$.
[M1] Take positive square root (since $y \ge 3$): $y - 3 = +\sqrt{x - 4} \implies y = 3 + \sqrt{x - 4}$.
[A1] $\mathbf{f^{-1}(x) = 3 + \sqrt{x - 4}}$ with domain $\mathbf{x \ge 4}$ (which corresponds to range of $f$).
Question 9 Tier 3: Extended 5 Marks

Given $f(x) = \dfrac{x}{x + 1}$ ($x \ne -1$). Define $f^1(x) = f(x)$, $f^2(x) = f(f(x))$, $f^3(x) = f(f^2(x))$, and in general $f^n(x) = f(f^{n-1}(x))$.
(a) Find simplified expressions for $f^2(x)$ and $f^3(x)$.
(b) Deduce a general conjecture for $f^n(x)$ in terms of $n$ and $x$.

View Full Method Markscheme
[M1] Compute $f^2(x) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + (x+1)} = \mathbf{\frac{x}{2x + 1}}$.
[M1] Compute $f^3(x) = f\left(\frac{x}{2x+1}\right) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + (2x+1)} = \mathbf{\frac{x}{3x + 1}}$.
[M1] Notice the pattern in the coefficients of $x$ in the denominator: for $n=1 \to 1$, $n=2 \to 2$, $n=3 \to 3$.
[A1] General conjecture: $\mathbf{f^n(x) = \dfrac{x}{nx + 1}}$.
[R1] Domain restriction: $x \ne -\frac{1}{k}$ for all $k \in \{1, 2, \dots, n\}$.
Question 10 Tier 3: Extended 5 Marks

Describe in exact mathematical sequence the geometric transformations needed to map the parent curve $y = x^2$ onto the graph of $y = 3(2x - 4)^2 + 5$.

View Full Method Markscheme
[M1] Factor out the coefficient inside the bracket: $2x - 4 = 2(x - 2)$. Thus $y = 3[2(x - 2)]^2 + 5$.
[M1] Horizontal dilation: A horizontal stretch by scale factor $\mathbf{\frac{1}{2}}$ parallel to the $x$-axis.
[M1] Horizontal translation: A translation of $\mathbf{2\text{ units to the right}}$ (vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$).
[M1] Vertical dilation: A vertical stretch by scale factor $\mathbf{3}$ parallel to the $y$-axis (or $3 \times 2^2 = 12$ if expanded as $12(x-2)^2+5$).
[A1] Vertical translation: A translation of $\mathbf{5\text{ units upwards}}$ (vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$).

Functions & Transformations Formulas

  • Function definition: For every $x \in \text{Domain}$, there exists a unique $y \in \text{Range}$.
  • Composite Function: $(f \circ g)(x) = f(g(x))$
  • Inverse Function Identity: $f(f^{-1}(x)) = f^{-1}(f(x)) = x$
  • Vertical Translation: $y = f(x) + k \implies$ Shift by $\begin{pmatrix} 0 \\ k \end{pmatrix}$
  • Horizontal Translation: $y = f(x - h) \implies$ Shift by $\begin{pmatrix} h \\ 0 \end{pmatrix}$
  • Reflection in $x$-axis: $y = -f(x)$
  • Reflection in $y$-axis: $y = f(-x)$
  • Vertical stretch: $y = a \cdot f(x)$ (scale factor $|a|$)
  • Horizontal stretch: $y = f(bx)$ (scale factor $\frac{1}{|b|}$)

Top Common MYP Pitfalls

1. Inverse Notation vs Reciprocal

$f^{-1}(x)$ denotes the inverse function, NOT the reciprocal $\frac{1}{f(x)}$. For reciprocals, write $[f(x)]^{-1}$.

2. Counter-Intuitive Horizontal Shifts

Thinking $f(x - 3)$ shifts left. Inside the argument, $(x - 3)$ shifts the graph right by $3$ units!

3. Order of Composite Functions

Evaluating $f(g(x))$ by substituting $f$ into $g$. Always evaluate from the inside out: find $g(x)$ first, then pass it as input to $f$.

4. Forgetting Horizontal Stretch Reciprocal

In $f(2x)$, stating the graph is stretched horizontally by factor $2$. It is actually compressed horizontally by factor $\frac{1}{2}$!

Interactive Mini Test: Functions & Transformations

Answer the questions to see your score and feedback.

Question 1 of 5 Function Domain

What is the domain of $f(x) = \dfrac{7}{\sqrt{x - 5}}$?

Question 2 of 5 Evaluating Composite Function

If $f(x) = 3x - 1$ and $g(x) = 2x^2$, what is $(g \circ f)(2)$?

Question 3 of 5 Inverse Function Formulation

Which function is the inverse of $f(x) = \dfrac{x + 3}{4}$?

Question 4 of 5 Horizontal Translation Direction

How is the graph of $y = f(x + 5)$ obtained from the graph of $y = f(x)$?

Question 5 of 5 Inverse Geometric Symmetry

The graph of $y = f^{-1}(x)$ is the geometric reflection of $y = f(x)$ across which line?