Functions and Transformations

Comprehensive guide to functions and transformations for IB Maths AA SL. Covers domain, range, inverses, composites, and graph transformations.

Functions are one of the most fundamental concepts in IB Mathematics Analysis and Approaches. They appear throughout the entire course — from algebra and calculus to statistics and modelling. A solid understanding of functions and how they transform is essential for success at both SL and HL.

In this guide, we cover the essential ideas: what a function is, how to describe its domain and range, how to find inverse and composite functions, and — crucially — how graph transformations work. Transformations are a favourite topic in IB exams because they test both algebraic manipulation and graphical understanding. The IB values conceptual understanding, so we will focus not just on how to perform these operations but why they work the way they do.

Your IB formula booklet contains the key transformation rules. The goal here is to understand the logic behind them so you can apply them confidently under exam conditions.

Core Concepts

What Is a Function?

A function ff is a rule that assigns to each input value (from the domain) exactly one output value (in the range). We write f:xf(x)f: x \mapsto f(x) or simply y=f(x)y = f(x).

The vertical line test is a quick visual check: if any vertical line intersects a graph more than once, the graph does not represent a function.

Domain and Range

The domain of a function is the set of all permitted input values (xx-values). The range is the set of all possible output values (yy-values).

Common domain restrictions include:

  • Division by zero: for f(x)=1x2f(x) = \frac{1}{x-2}, we need x2x \neq 2
  • Square roots of negatives: for f(x)=x+3f(x) = \sqrt{x+3}, we need x3x \geq -3
  • Logarithms of non-positive numbers: for f(x)=ln(x)f(x) = \ln(x), we need x>0x > 0

To find the range, consider the behaviour of the function — its maximum and minimum values, asymptotes, and end behaviour. Sketching the graph (by hand or on your GDC) is often the most efficient approach.

Composite Functions

The composite function (fg)(x)=f(g(x))(f \circ g)(x) = f(g(x)) means "apply gg first, then ff." The order matters: f(g(x))g(f(x))f(g(x)) \neq g(f(x)) in general.

To evaluate f(g(x))f(g(x)), replace every xx in the formula for ff with the expression g(x)g(x).

Example: If f(x)=2x+1f(x) = 2x + 1 and g(x)=x2g(x) = x^2, then:

  • f(g(x))=f(x2)=2x2+1f(g(x)) = f(x^2) = 2x^2 + 1
  • g(f(x))=g(2x+1)=(2x+1)2=4x2+4x+1g(f(x)) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1

The domain of fgf \circ g is the set of xx-values in the domain of gg for which g(x)g(x) is in the domain of ff.

Inverse Functions

The inverse function f1f^{-1} reverses the effect of ff. If f(a)=bf(a) = b, then f1(b)=af^{-1}(b) = a. Formally, f(f1(x))=f1(f(x))=xf(f^{-1}(x)) = f^{-1}(f(x)) = x.

A function has an inverse if and only if it is one-to-one (injective) — each output comes from exactly one input. Graphically, this means it passes the horizontal line test.

To find the inverse algebraically:

  1. Write y=f(x)y = f(x)
  2. Swap xx and yy
  3. Solve for yy
  4. Write f1(x)=yf^{-1}(x) = y

Graphical property: The graph of f1f^{-1} is the reflection of the graph of ff in the line y=xy = x.

Key relationship: The domain of f1f^{-1} is the range of ff, and the range of f1f^{-1} is the domain of ff.

Self-Inverse Functions

A function is self-inverse if f1(x)=f(x)f^{-1}(x) = f(x), which means f(f(x))=xf(f(x)) = x. Examples include f(x)=1xf(x) = \frac{1}{x} and f(x)=xf(x) = -x. The graph of a self-inverse function is symmetric about the line y=xy = x.

Graph Transformations

Transformations modify the graph of a function. The IB formula booklet includes these, but understanding the logic is key.

Translations (Shifts)

Transformation Effect on graph Example
y=f(x)+ay = f(x) + a Vertical shift up by aa units y=x2+3y = x^2 + 3 shifts parabola up 3
y=f(x)ay = f(x) - a Vertical shift down by aa units y=x22y = x^2 - 2 shifts parabola down 2
y=f(xa)y = f(x - a) Horizontal shift right by aa units y=(x1)2y = (x-1)^2 shifts right 1
y=f(x+a)y = f(x + a) Horizontal shift left by aa units y=(x+4)2y = (x+4)^2 shifts left 4

The crucial insight: horizontal transformations are "opposite to what you might expect." The expression f(x3)f(x - 3) shifts the graph right, not left. Think of it this way: to get the same yy-value, xx must be 3 units larger.

Using vector notation, y=f(xa)+by = f(x-a) + b is a translation by the vector (ab)\begin{pmatrix} a \\ b \end{pmatrix}.

Reflections

Transformation Effect on graph
y=f(x)y = -f(x) Reflection in the xx-axis (multiply all yy-values by 1-1)
y=f(x)y = f(-x) Reflection in the yy-axis (replace xx with x-x)

Stretches (Dilations)

Transformation Effect on graph
y=af(x)y = af(x) Vertical stretch by factor aa (multiply all yy-values by aa)
y=f(bx)y = f(bx) Horizontal stretch by factor 1b\frac{1}{b} (compress by factor bb)

Again, horizontal stretches are "opposite": y=f(2x)y = f(2x) compresses the graph horizontally by a factor of 2 (everything happens twice as fast).

Order of Transformations

When multiple transformations are combined, the order matters. A useful approach is to think of the transformation as acting on the xx or yy variable:

  1. Inside the function (affecting xx): Apply in reverse order — stretches first, then translations.
  2. Outside the function (affecting yy): Apply in the written order — stretches first, then translations.

For example, y=2f(x3)+1y = 2f(x - 3) + 1:

  • Horizontal shift right 3 (inside)
  • Vertical stretch by factor 2 (outside)
  • Vertical shift up 1 (outside)

Strategy Tips

Tip 1: Sketch Graphs on Your GDC

When unsure about a transformation, graph both the original and transformed function on your GDC to visually confirm the effect. This builds intuition and catches algebraic errors.

Tip 2: Track Key Points Through Transformations

Choose 3-4 key points on the original graph (intercepts, vertex, endpoints). Apply the transformation to each point. Plot the new points and connect them. This method is reliable and scores well in exams.

Tip 3: Remember the "Opposite" Rule for Horizontal Changes

Anything that happens inside the function (affecting xx) works in the opposite direction to what you might expect. f(x3)f(x-3) shifts right, f(2x)f(2x) compresses. Train yourself to pause and think about horizontal transformations.

Tip 4: Use the Swap Method for Inverses

The "swap xx and yy, then solve" method is the most reliable way to find inverse functions. Always verify by checking that f(f1(x))=xf(f^{-1}(x)) = x.

Tip 5: State Domain Restrictions Explicitly

In IB exams, marks are often awarded for stating the domain and range. Do not skip this step, especially for inverse functions where the domain of f1f^{-1} equals the range of ff.

Worked Example: Example 1

Problem

Let f(x)=3x2f(x) = 3x - 2 and g(x)=x2+1g(x) = x^2 + 1. Find: (a) (fg)(x)(f \circ g)(x) (b) (gf)(2)(g \circ f)(2)

Solution

(a) (fg)(x)=f(g(x))=f(x2+1)=3(x2+1)2=3x2+32=3x2+1(f \circ g)(x) = f(g(x)) = f(x^2 + 1) = 3(x^2 + 1) - 2 = 3x^2 + 3 - 2 = 3x^2 + 1

(b) First: f(2)=3(2)2=4f(2) = 3(2) - 2 = 4. Then: g(4)=42+1=17g(4) = 4^2 + 1 = 17.

So (gf)(2)=17(g \circ f)(2) = 17.

Worked Example: Example 2

Problem

Find the inverse of f(x)=2x+1x3f(x) = \frac{2x+1}{x-3}, x3x \neq 3, and state its domain.

Solution

Let y=2x+1x3y = \frac{2x+1}{x-3}.

Swap xx and yy: x=2y+1y3x = \frac{2y+1}{y-3}.

Solve for yy: x(y3)=2y+1x(y-3) = 2y + 1

xy3x=2y+1xy - 3x = 2y + 1

xy2y=3x+1xy - 2y = 3x + 1

y(x2)=3x+1y(x - 2) = 3x + 1

y=3x+1x2y = \frac{3x+1}{x-2}

So f1(x)=3x+1x2f^{-1}(x) = \frac{3x+1}{x-2}, with domain x2x \neq 2.

Note: The domain of f1f^{-1} (x2x \neq 2) corresponds to the range of ff (which is all real numbers except 2, the value the horizontal asymptote approaches).

Worked Example: Example 3

Problem

The graph of y=f(x)y = f(x) passes through the point (4,6)(4, 6). Find the corresponding point on the graph of y=3f(2x1)+4y = 3f(2x - 1) + 4.

Solution

We need 2x1=42x - 1 = 4, so x=2.5x = 2.5.

When x=2.5x = 2.5: y=3f(2(2.5)1)+4=3f(4)+4=3(6)+4=22y = 3f(2(2.5)-1) + 4 = 3f(4) + 4 = 3(6) + 4 = 22.

The corresponding point is (2.5,22)(2.5, 22).

Worked Example: Example 4

Problem

Describe a sequence of transformations that maps y=x2y = x^2 to y=(x+2)2+5y = -(x+2)^2 + 5.

Solution

Comparing y=x2y = x^2 with y=(x+2)2+5y = -(x+2)^2 + 5:

  1. Reflection in the xx-axis: y=x2y=x2y = x^2 \to y = -x^2
  2. Translation left 2 units: y=x2y=(x+2)2y = -x^2 \to y = -(x+2)^2
  3. Translation up 5 units: y=(x+2)2y=(x+2)2+5y = -(x+2)^2 \to y = -(x+2)^2 + 5

Using vector notation, after the reflection, the graph is translated by (25)\begin{pmatrix} -2 \\ 5 \end{pmatrix}.

Practice Problems

  1. Problem 1

    Let f(x)=1x+1f(x) = \frac{1}{x+1}, x1x \neq -1. Find f1(x)f^{-1}(x) and verify that f(f1(x))=xf(f^{-1}(x)) = x.

    Problem 2

    The graph of y=f(x)y = f(x) has a vertex at (3,2)(3, -2). Find the coordinates of the vertex after the transformation y=2f(x1)+3y = 2f(x-1) + 3.

    Problem 3

    Given f(x)=2x5f(x) = 2x - 5 and g(x)=x+52g(x) = \frac{x+5}{2}, show that ff and gg are inverse functions.

    Problem 4

    Sketch the graph of y=f(x)y = |f(x)| given that f(x)=x24x+3f(x) = x^2 - 4x + 3. State the range of y=f(x)y = |f(x)|.

    Problem 5

    The function hh is defined by h(x)=4x+1h(x) = \frac{4}{x} + 1 for x>0x > 0. Show that hh is self-inverse.

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Common Mistakes

  • Getting the direction of horizontal translations wrong. Remember: f(xa)f(x - a) shifts right by aa, and f(x+a)f(x + a) shifts left by aa. This is the single most common error in transformation questions.

  • Confusing the order in composite functions. f(g(x))f(g(x)) means apply gg first, then ff. Read the notation carefully: fgf \circ g means "ff after gg."

  • Forgetting to swap domain and range for inverses. The domain of f1f^{-1} is the range of ff. If you restrict the domain of ff (e.g., to make it one-to-one), this determines the range of f1f^{-1}.

  • Applying transformations in the wrong order. When combining multiple transformations, the order of operations matters. Horizontal changes (inside the function) act in reverse order.

  • Not stating the domain. Many IB mark schemes award a specific mark for stating the domain of the inverse function. Never skip this.

  • Assuming all functions have inverses. A function must be one-to-one to have an inverse. If it is not (e.g., f(x)=x2f(x) = x^2 on all of R\mathbb{R}), you must restrict the domain first.

Frequently Asked Questions

Why do horizontal transformations work in the opposite direction?

Think about it this way: for y=f(x3)y = f(x - 3), the function outputs the same yy-value as ff, but the input needs to be 3 units larger to compensate for the subtraction. So the graph shifts right. The key is that the transformation acts on the input, not the output.

Do I need to memorise all the transformation rules?

The transformation rules are summarised in your IB formula booklet. However, you should understand them well enough that you can reconstruct them quickly. Exam questions require fluent application, not just recall.

How do I find the domain and range of a composite function?

The domain of f(g(x))f(g(x)) is all values of xx in the domain of gg such that g(x)g(x) is in the domain of ff. The range can be found by considering what outputs gg produces and how ff acts on those values. Graphing on your GDC is often the fastest way.

What is the difference between $f^{-1}(x)$ and $\frac{1}{f(x)}$?

f1(x)f^{-1}(x) is the inverse function (it reverses the mapping). 1f(x)=[f(x)]1\frac{1}{f(x)} = [f(x)]^{-1} is the reciprocal of the function value. These are completely different! For example, if f(x)=2xf(x) = 2x, then f1(x)=x2f^{-1}(x) = \frac{x}{2}, but 1f(x)=12x\frac{1}{f(x)} = \frac{1}{2x}.

How are transformations tested in the IB exam?

Transformations appear in multiple ways: describing transformations given an equation, writing the equation from a description, finding images of points, and sketching transformed graphs. You should be comfortable with all these question types.

Key Takeaways

  • Functions assign exactly one output to each input. The domain is the set of valid inputs, and the range is the set of possible outputs.

  • Composite functions apply one function after another. In f(g(x))f(g(x)), apply gg first, then ff. Order matters.

  • Inverse functions reverse the mapping. Use the swap-and-solve method. Remember that the domain and range swap between a function and its inverse.

  • Horizontal transformations are counterintuitive. Changes inside the function brackets (xx-direction) act in the opposite direction. This is the most important thing to internalise.

  • Track key points through transformations. This is the most reliable method for sketching transformed graphs and is a standard IB approach.

  • Always state the domain. Especially for inverse functions and compositions, the domain is worth marks and shows mathematical precision.

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