Functions (Advanced)

Prerequisites

This note assumes the reader is already familiar with:

• Basic functions: notation, domain, range, evaluation, and graphing.
• Linear and quadratic functions: slopes, intercepts, vertex form, factorisation.
• Simple graph transformations: translation, reflection, and scaling of $y = f(x)$.

These topics are covered in the introductory Functions note. Readers who have completed the DSE compulsory mathematics core should proceed directly; others should review the introductory functions material first.

This note extends the treatment of functions covered in Functions, focusing on Domain restrictions, composite and inverse functions with non-trivial domains, and graphical Transformations.

Domain and Range

Natural Domain

The natural domain of a function is the largest subset of $\mathbb{R}$ for which the function Expression is defined. Restrictions arise from:

Finding the Range

To find the range of $f(x)$:

1. Complete the square (for quadratics).
2. Consider the behaviour of the function at critical points and at the boundaries of the domain.
3. For rational functions, find horizontal asymptotes and analyse sign changes.

Worked Example 1

Find the domain and range of $f(x) = \sqrt{4 - x^2}$.

Domain: $4 - x^2 \geqslant 0 \implies x^2 \leqslant 4 \implies -2 \leqslant x \leqslant 2$.

Range: Since $4 - x^2$ ranges from $0$ (at $x = \pm 2$) to $4$ (at $x = 0$), and $\sqrt{\cdot}$ is Non-negative: $\mathrm{range}(f) = [0, 2]$.

Composite Functions

Definition

Given $f$ and $g$The composite $f \circ g$ is:

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

Domain of a Composite

$\mathrm{dom}(f \circ g) = \{x \in \mathrm{dom}(g) : g(x) \in \mathrm{dom}(f)\}$

Worked Example 2

Let $f(x) = \sqrt{x + 1}$ and $g(x) = x^2 - 4$. Find $\mathrm{dom}(f \circ g)$.

$\mathrm{dom}(g) = \mathbb{R}$.

$\mathrm{dom}(f) = [ -1, \infty)$So we need $g(x) \geqslant -1$I.e., $x^2 - 4 \geqslant -1 \implies x^2 \geqslant 3$.

$\mathrm{dom}(f \circ g) = (-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$

Worked Example 3

Let $f(x) = \dfrac{1}{x}$ and $g(x) = x + 1$. Find $f \circ g$, $g \circ f$And their domains.

$(f \circ g)(x) = f(g(x)) = f(x + 1) = \dfrac{1}{x + 1}$ $\mathrm{dom} = \mathbb{R} \setminus \{-1\}$.

$(g \circ f)(x) = g(f(x)) = g\!\left(\dfrac{1}{x}\right) = \dfrac{1}{x} + 1$ $\mathrm{dom} = \mathbb{R} \setminus \{0\}$.

Inverse Functions

Existence Condition

A function $f$ has an inverse $f^{-1}$ if and only if $f$ is bijective (one-to-one and onto). If the Natural domain of $f$ does not yield injectivity, restrict the domain.

Procedure to Find $f^{-1}$

1. Set $y = f(x)$.
2. Solve for $x$ in terms of $y$.
3. Interchange $x$ and $y$ to obtain $f^{-1}(x)$.

The domain of $f^{-1}$ equals the range of $f$And vice versa.

Graphical Relationship

The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$.

Worked Example 4

Find the inverse of $f(x) = \dfrac{2x - 3}{x + 1}$ for $x \neq -1$.

Set $y = \dfrac{2x - 3}{x + 1}$.

$y(x + 1) = 2x - 3 \implies yx + y = 2x - 3 \implies yx - 2x = -3 - y \implies x(y - 2) = -(y + 3)$

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

Therefore $f^{-1}(x) = \dfrac{x + 3}{2 - x}$With domain $\mathbb{R} \setminus \{2\}$.

Function Transformations

Individual Transformations

Given $y = f(x)$:

| Transformation | Effect on Graph | | -------------- | ------------------------------------------------------------- | --- | ----------------------------------- | | $y = f(x) + c$ | Vertical shift up by $c$ ($c \gt 0$) or down ($c \lt 0$) | | | | $y = f(x - h)$ | Horizontal shift right by $h$ ($h \gt 0$) or left ($h \lt 0$) | | | | $y = af(x)$ | Vertical stretch by factor $| a |$; reflect in $x$-axis if $a \lt 0$ | | $y = f(kx)$ | Horizontal stretch by factor $1/ | k |$; reflect in $y$-axis if $k \lt 0$ |

Combined Transformation: $y = af(x + b) + c$

Apply in order from inside out:

1. Horizontal shift by $-b$
2. Vertical stretch/reflection by factor $a$
3. Vertical shift by $c$

Worked Example 5

The graph of $y = f(x)$ passes through $(2, 5)$ and $(4, -1)$. Find the corresponding points on $y = -2f(x - 3) + 1$.

For $(2, 5)$: set $x - 3 = 2 \implies x = 5$. Then $y = -2(5) + 1 = -9$. Point: $(5, -9)$.

For $(4, -1)$: set $x - 3 = 4 \implies x = 7$. Then $y = -2(-1) + 1 = 3$. Point: $(7, 3)$.

Worked Example 6

Describe the transformation from $y = \sqrt{x}$ to $y = \sqrt{3 - x} + 2$.

$y = \sqrt{-(x - 3)} + 2 = f(-(x - 3)) + 2$ where $f(x) = \sqrt{x}$.

1. Reflect in $y$-axis: $y = \sqrt{-x}$
2. Shift right by 3: $y = \sqrt{-(x - 3)} = \sqrt{3 - x}$
3. Shift up by 2: $y = \sqrt{3 - x} + 2$

Domain: $3 - x \geqslant 0 \implies x \leqslant 3$. Range: $[2, \infty)$.

Piecewise Functions

A piecewise function is defined by different expressions on different intervals of its domain.

Worked Example 7

$f(x) = \begin{cases} x^2 & \mathrm{if } x \lt 0 \\ 2x + 1 & \mathrm{if } 0 \leqslant x \leqslant 3 \\ 10 - x & \mathrm{if } x \gt 3 \end{cases}$

Find $f(-2)$, $f(0)$, $f(3)$And $f(5)$.

f(-2) = (-2)^2 = 4$$f(0) = 2(0) + 1 = 1$$f(3) = 2(3) + 1 = 7$$f(5) = 10 - 5 = 5.

Common Pitfalls

• When finding the domain of $f \circ g$Applying the domain restrictions of $f$ to $x$ instead of to $g(x)$. The argument of $f$ must be valid, so it is $g(x)$ that must fall in $\mathrm{dom}(f)$.
• Forgetting that $f \circ g \neq g \circ f$ . Always check the order.
• When finding an inverse, forgetting to verify that the function is one-to-one on the given domain.
• Confusing $y = f(-x)$ (reflection in $y$-axis) with $y = -f(x)$ (reflection in $x$-axis).
• For piecewise functions, using the wrong expression for a given $x$-value.

Summary Table

Additional Worked Examples

Worked Example 8: Domain of a composite with square root and rational function

Let $f(x) = \dfrac{x + 1}{x - 2}$ and $g(x) = \sqrt{x - 3}$. Find $\mathrm{dom}(f \circ g)$ and $\mathrm{dom}(g \circ f)$.

Worked Example 9: Inverse of a restricted quadratic

Let $f(x) = 2x^2 - 8x + 5$ with domain $[2, \infty)$. Find $f^{-1}$ and state its domain and range.

Worked Example 10: Transformation of multiple points

The graph of $y = f(x)$ passes through $(1, 4)$ and has a local minimum at $(2, -1)$. Find the corresponding points on $y = 2f(3x - 6) + 5$.

Worked Example 11: Composite with logarithm

Let $f(x) = \ln(x - 1)$ and $g(x) = x^2 + 1$. Find $(f \circ g)(x)$, $(g \circ f)(x)$And their domains.

Worked Example 12: Self-inverse function

Show that $f(x) = \dfrac{3x - 2}{x - 3}$ ($x \neq 3$) is self-inverse.

Additional Common Pitfalls

1. Applying domain restrictions to $x$ instead of $g(x)$. When finding $\mathrm{dom}(f \circ g)$The condition $g(x) \in \mathrm{dom}(f)$ must be applied to the expression $g(x)$Not to $x$ directly. Always substitute first, then impose domain conditions.

2. Assuming injectivity on the natural domain. A quadratic $ax^2 + bx + c$ is only one-to-one on a half-domain $(-\infty,\; -b/(2a)]$ or $[-b/(2a),\; \infty)$. Before finding an inverse, verify or restrict the domain.

3. Choosing the wrong branch of the inverse. When $f(x) = x^2$ is restricted to $(-\infty, 0]$The inverse is $f^{-1}(x) = -\sqrt{x}$Not $+\sqrt{x}$. Always match the sign to the restricted domain.

4. Composition order confusion. $(f \circ g)(x) = f(g(x))$ means $g$ is applied first, then $f$. The notation reads right-to-left: $(f \circ g)(x)$ is ”$f$ of $g$ of $x$”.

5. Transformation order errors. For $y = af(kx + b) + c$Apply from inside out: horizontal shift by $-b$Horizontal stretch by $1/k$Vertical stretch by $a$Vertical shift by $c$. Mixing up this order is a very common mistake.

6. Ignoring the range when checking invertibility. Even if $f$ is one-to-one on its domain, the codomain must equal the range for $f$ to be bijective. In DSE problems, the codomain is assumed to be the range unless stated otherwise.

7. Forgetting that $f \circ g$ and $g \circ f$ differ. , $f \circ g \neq g \circ f$. Always compute each separately and check domains independently.

8. Piecewise function boundary values. At the boundary between two pieces, always check which expression applies. If the definition uses $\leq$ for one piece and $\lt$ for the next, the boundary point belongs to the $\leq$ piece.

Exam-Style Problems

Problem 1. Let $f(x) = \dfrac{2x + 3}{x - 1}$ ($x \neq 1$) and $g(x) = \sqrt{x + 2}$. Find $(f \circ g)(x)$ and its domain.

Problem 2. Find the inverse of $f(x) = \dfrac{2x - 1}{x + 3}$ ($x \neq -3$). Hence determine whether $f(x) = f^{-1}(x)$ has any real solutions.

Problem 3. The function $f$ is defined by $f(x) = x^2 + 4x$ for $x \geq -2$. Find $f^{-1}(5)$.

Problem 4. Describe fully the sequence of transformations mapping $y = x^2$ to $y = 2(3-x)^2 + 1$.

Problem 5. Let $f(x) = \dfrac{1}{x+1}$ ($x \neq -1$) and $g(x) = x^2$. Find $(f \circ g \circ f)(x)$ and its domain.

Problem 6. Given $f(x) = |2x - 1| + |x + 3|$Find the minimum value of $f$.

Problem 7. If $f(x) = \dfrac{x}{x^2 + 1}$Find the range of $f$.

Problem 8. If $f(x) = 2x - 1$ and $g(x) = x + 3$Find the linear function $h(x)$ such that $(f \circ h)(x) = (g \circ f)(x)$ for all $x$.

Cross-References

• Basic Functions: Foundational definitions and notation are in Functions.
• Quadratics: Quadratic functions feature heavily in inverse function problems. See Quadratics.
• Inequalities: Domain restrictions often involve solving inequalities. See the inequalities notes.
• Coordinate Geometry: Graphical interpretations of functions and transformations. See Coordinate Geometry.

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DSE Exam Technique

Showing Working

For function problems in DSE Paper 1:

1. When finding the domain of a composite function, explicitly state $\mathrm{dom}(g)$ and the condition $g(x) \in \mathrm{dom}(f)$.
2. When finding an inverse, write $y = f(x)$Solve for $x$And then interchange.
3. When checking invertibility, verify that the function is one-to-one (strictly increasing or decreasing on the domain).
4. For transformation problems, identify the sequence of transformations from inside out.

Significant Figures

Coordinate answers should be exact where possible. Decimal answers to 3 significant figures.

Common DSE Question Types

1. Domain of composite functions (especially with square roots and rational functions).
2. Finding inverse functions (restricted quadratics, rational functions).
3. Transformation of points (tracking specific points through a series of transformations).
4. Self-inverse verification (show $f^{-1} = f$).
5. Range finding (using the discriminant method for rational functions).

Additional Worked Examples

Worked Example 13: Domain with nested functions

Let $f(x) = \dfrac{1}{\sqrt{x - 2}}$ and $g(x) = x^2 + 1$. Find $\mathrm{dom}(f \circ g)$ and $\mathrm{dom}(g \circ f)$.

Worked Example 14: Inverse of a restricted rational function

Let $f(x) = \dfrac{2x}{x - 3}$ for $x > 3$. Find $f^{-1}$.

Worked Example 15: Even and odd functions

Determine whether $f(x) = \dfrac{x}{x^2 + 1}$ is even, odd, or neither.