Functions (Advanced)

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.

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

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.

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: $\mathrm{range}(f) = [0, 2]$.

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

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., $x^2 - 4 \geqslant -1 \implies x^2 \geqslant 3$.

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

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

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

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.

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.

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

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

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

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

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

Given $y = f(x)$:

Apply in order from inside out:

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

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: $(7, 3)$.

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: $[2, \infty)$.

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

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.

• 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.

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.

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.

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 $x$.

• 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.

• Linear Algebra: Vector Spaces, Generalisation of functions to vector-space morphisms.
• Real Analysis: Continuity, Rigorous epsilon-delta definition of the continuity studied informally here.

Behind every scientific discovery and technological innovation lies mathematics. Functions model relationships between variables, statistics reveals patterns in data, and logic ensures rigorous reasoning. Mathematics teaches us to think precisely, solve systematically, and communicate evidently - skills that are valuable far beyond the classroom.