Functions (Advanced)

This note extends the treatment of functions covered in functions.md), 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:

Restriction	Condition	Example
Division by zero	Denominator $\neq 0$	$f(x) = \dfrac{1}{x - 2}$: $\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'} \setminus \{2\}$
Even root	Radicand $\geqslant 0$	$f(x) = \sqrt{x - 3}$: $\mathrm{dom}(f) = [3, \infty)$
Logarithm	Argument $\gt 0$	$f(x) = \ln(x + 1)$: $\mathrm{dom}(f) = (-1, \infty)$

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

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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.

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

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

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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 $\mathbb{'\{'}R{'\}'} \setminus \{2\}$.

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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: $[2, \infty)$.

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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 $f(5)$.

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

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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 $f \circ g \neq g \circ f$ in general. 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.

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Summary Table

Topic	Key Result
Domain of $f \circ g$	$\{x \in \mathrm{dom}(g) : g(x) \in \mathrm{dom}(f)\}$
Inverse existence	$f$ must be bijective
$f^{-1}(f(x)) = x$	For all $x \in \mathrm{dom}(f)$
$f(f^{-1}(x)) = x$	For all $x \in \mathrm{dom}(f^{-1})$
Graph of inverse	Reflection in $y = x$
$y = f(x - h)$	Shift right by $h$

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Wrap-up Questions

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

$(f \circ g)(x) = f(2x - 3) = \dfrac{2x - 3 + 2}{2x - 3 - 1} = \dfrac{2x - 1}{2x - 4}$.

$\mathrm{dom}(g) = \mathbb{'\{'}R{'\}'}$. $\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'} \setminus \{1\}$, so $g(x) \neq 1$: $2x - 3 \neq 1 \implies x \neq 2$. Also $2x - 4 \neq 0 \implies x \neq 2$. $\mathrm{dom}(f \circ g) = \mathbb{'\{'}R{'\}'} \setminus \{2\}$.

2. Question: Find $f^{-1}$ for $f(x) = \dfrac{3x + 1}{x - 2}$ ($x \neq 2$).

Set $y = \dfrac{3x + 1}{x - 2}$. Then $y(x - 2) = 3x + 1 \implies yx - 2y = 3x + 1 \implies x(y - 3) = 2y + 1$.

$f^{-1}(x) = \dfrac{2x + 1}{x - 3}$, $\mathrm{dom}(f^{-1}) = \mathbb{'\{'}R{'\}'} \setminus \{3\}$.

3. Question: Let $f(x) = x^2 - 4x + 3$ with domain $[1, \infty)$. Find $f^{-1}(0)$.

First find $f^{-1}$. Set $y = (x - 2)^2 - 1$. Since domain is $[1, \infty)$, range is $[-1, \infty)$.

$(x - 2)^2 = y + 1 \implies x - 2 = \sqrt{y + 1}$ (positive root since $x \geqslant 1$).

$f^{-1}(x) = 2 + \sqrt{x + 1}$, $\mathrm{dom}(f^{-1}) = [-1, \infty)$.

$f^{-1}(0) = 2 + \sqrt{0 + 1} = 2 + 1 = 3$.

Verification: $f(3) = 9 - 12 + 3 = 0$. Confirmed.

4. Question: The graph of $y = f(x)$ has a minimum at $(1, -2)$ and passes through $(0, 3)$. Find the corresponding points on $y = 3f(2x) + 1$.

$(1, -2) \to$ set $2x = 1 \implies x = 0.5$, $y = 3(-2) + 1 = -5$. Point: $(0.5, -5)$.

$(0, 3) \to$ set $2x = 0 \implies x = 0$, $y = 3(3) + 1 = 10$. Point: $(0, 10)$.

5. Question: A function $f$ is defined by $f(x) = 2 - x^2$ for $x \leqslant 0$. State the range of $f$ and find $f^{-1}$.

Since $x \leqslant 0$: $x^2 \geqslant 0$, so $f(x) = 2 - x^2 \leqslant 2$. As $x \to -\infty$, $f(x) \to -\infty$. Range: $(-\infty, 2]$.

Set $y = 2 - x^2 \implies x^2 = 2 - y \implies x = -\sqrt{2 - y}$ (negative root since $x \leqslant 0$).

$f^{-1}(x) = -\sqrt{2 - x}$, $\mathrm{dom}(f^{-1}) = (-\infty, 2]$.

6. Question: Given $f(x) = \sqrt{x - 1}$ and $g(x) = x^2 + x + 1$, find $\mathrm{dom}(g \circ f)$.

$\mathrm{dom}(f) = [1, \infty)$, $\mathrm{dom}(g) = \mathbb{'\{'}R{'\}'}$.

$(g \circ f)(x) = g(\sqrt{x - 1}) = (\sqrt{x - 1})^2 + \sqrt{x - 1} + 1 = x - 1 + \sqrt{x - 1} + 1 = x + \sqrt{x - 1}$.

Since $g$ has no domain restriction, $\mathrm{dom}(g \circ f) = \mathrm{dom}(f) = [1, \infty)$.

7. Question: Find the domain and range of $f(x) = \dfrac{1}{x^2 + 1}$.

Domain: $x^2 + 1 \neq 0$ for all real $x$ (since $x^2 \geqslant 0$). $\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'}$.

Range: $x^2 + 1 \geqslant 1$, so $0 \lt \dfrac{1}{x^2 + 1} \leqslant 1$. $\mathrm{range}(f) = (0, 1]$.

8. Question: Let $f(x) = |x - 3| + |x + 1|$. Express $f$ as a piecewise function and find its minimum value.

Critical points at $x = 3$ and $x = -1$:

$f(x) = \begin{cases} -(x - 3) + -(x + 1) = -2x + 2 & \mathrm{if } x \lt -1 \\ -(x - 3) + (x + 1) = 4 & \mathrm{if } -1 \leqslant x \leqslant 3 \\ (x - 3) + (x + 1) = 2x - 2 & \mathrm{if } x \gt 3 \end{cases}$

For $x \lt -1$: $f(x) = -2x + 2$, which is decreasing (as $x$ increases towards $-1$). As $x \to -1^-$: $f(x) \to 4$.

For $-1 \leqslant x \leqslant 3$: $f(x) = 4$ (constant).

For $x \gt 3$: $f(x) = 2x - 2$, which is increasing.

Minimum value: $4$, attained for all $x \in [-1, 3]$.

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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 $\mathrm{dom}(g \circ f)$.

Solution

$\mathrm{dom}(g) = [3, \infty)$. $\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'} \setminus \{2\}$.

For $f \circ g$: We need $g(x) \in \mathrm{dom}(f)$, i.e., $\sqrt{x-3} \neq 2$.

$\sqrt{x-3} = 2 \implies x = 7$. So exclude $x = 7$.

$\mathrm{dom}(f \circ g) = [3, \infty) \setminus \{7\}$.

For $g \circ f$: We need $x \in \mathrm{dom}(f)$ and $f(x) \geq 3$.

$\frac{x+1}{x-2} \geq 3 \implies \frac{x+1-3(x-2)}{x-2} \geq 0 \implies \frac{-2x+7}{x-2} \geq 0$

Critical points: $x = \dfrac{7}{2}$ and $x = 2$.

Interval	Sign of $\dfrac{-2x+7}{x-2}$
$x \lt 2$	negative / negative $=$ positive
$2 \lt x \lt \dfrac{7}{2}$	positive / positive $=$ positive
$x \gt \dfrac{7}{2}$	negative / positive $=$ negative

At $x = \dfrac{7}{2}$: expression equals $0$, which satisfies $\geq 0$.

$\mathrm{dom}(g \circ f) = \left(2,\; \dfrac{7}{2}\right]$.

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.

Solution

Complete the square: $f(x) = 2(x^2 - 4x) + 5 = 2\!\left[(x-2)^2 - 4\right] + 5 = 2(x-2)^2 - 3$.

Since the domain is $[2, \infty)$ and the vertex is at $x = 2$, the function is strictly increasing and hence one-to-one.

Range: $[-3, \infty)$.

Set $y = 2(x-2)^2 - 3$:

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

$x = 2 + \sqrt{\frac{y+3}{2}}$

(positive root since $x \geq 2$)

$f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}}$

$\mathrm{dom}(f^{-1}) = [-3, \infty)$, $\mathrm{range}(f^{-1}) = [2, \infty)$.

Verification: $f^{-1}(f(3)) = f^{-1}(-1) = 2 + \sqrt{1} = 3$. Correct.

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 $y = 2f(3x - 6) + 5$.

Solution

Rewrite: $y = 2f(3(x-2)) + 5$.

For $(1, 4)$ on $y = f(x)$: Set $3(x-2) = 1$, so $x - 2 = \dfrac{1}{3}$, giving $x = \dfrac{7}{3}$.

$y = 2(4) + 5 = 13$

Corresponding point: $\left(\dfrac{7}{3},\; 13\right)$.

For the minimum at $(2, -1)$: Set $3(x-2) = 2$, so $x - 2 = \dfrac{2}{3}$, giving $x = \dfrac{8}{3}$.

$y = 2(-1) + 5 = 3$

Corresponding point: $\left(\dfrac{8}{3},\; 3\right)$. This is the minimum of the transformed graph.

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.

Solution

$\mathrm{dom}(g) = \mathbb{'\{'}R{'\}'}$, $\mathrm{dom}(f) = (1, \infty)$.

$(f \circ g)(x) = f(g(x)) = \ln(x^2 + 1 - 1) = \ln(x^2)$.

Domain: need $g(x) \in \mathrm{dom}(f)$, i.e., $x^2 + 1 \gt 1 \implies x^2 \gt 0 \implies x \neq 0$.

$\mathrm{dom}(f \circ g) = \mathbb{'\{'}R{'\}'} \setminus \{0\}$.

$(g \circ f)(x) = g(f(x)) = [\ln(x-1)]^2 + 1$.

Domain: $\mathrm{dom}(g \circ f) = \mathrm{dom}(f) = (1, \infty)$.

Worked Example 12: Self-inverse function

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

Solution

Set $y = \dfrac{3x - 2}{x - 3}$:

$y(x - 3) = 3x - 2 \implies xy - 3y = 3x - 2 \implies xy - 3x = 3y - 2$

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

Interchanging $x$ and $y$:

$f^{-1}(x) = \frac{3x - 2}{x - 3} = f(x)$

Since $f^{-1} = f$, the function is self-inverse.

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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 usually assumed to be the range unless stated otherwise.

7. Forgetting that $f \circ g$ and $g \circ f$ differ. In general, $f \circ g \neq g \circ f$. 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.

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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.

Solution

$(f \circ g)(x) = f(\sqrt{x+2}) = \frac{2\sqrt{x+2} + 3}{\sqrt{x+2} - 1}$

$\mathrm{dom}(g) = [-2, \infty)$. $\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'} \setminus \{1\}$.

Need $\sqrt{x+2} \neq 1 \implies x + 2 \neq 1 \implies x \neq -1$.

$\mathrm{dom}(f \circ g) = [-2, -1) \cup (-1, \infty)$.

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.

Solution

Set $y = \dfrac{2x-1}{x+3}$: $y(x+3) = 2x-1 \implies xy + 3y = 2x - 1 \implies x(y-2) = -1 - 3y$.

$f^{-1}(x) = \frac{-1-3x}{x-2} = \frac{3x+1}{2-x}$

For $f(x) = f^{-1}(x)$:

$\frac{2x-1}{x+3} = \frac{3x+1}{2-x}$

$(2x-1)(2-x) = (3x+1)(x+3)$

$4x - 2x^2 - 2 + x = 3x^2 + 9x + x + 3$

$-2x^2 + 5x - 2 = 3x^2 + 10x + 3$

$-5x^2 - 5x - 5 = 0 \implies x^2 + x + 1 = 0$

$\Delta = 1 - 4 = -3 \lt 0$. No real solutions.

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

Solution

$f(x) = (x+2)^2 - 4$. Since $x \geq -2$ and the vertex is at $x = -2$, $f$ is strictly increasing.

Range: $[-4, \infty)$. Since $5 \geq -4$, $f^{-1}(5)$ exists.

Set $(x+2)^2 - 4 = 5 \implies (x+2)^2 = 9 \implies x + 2 = 3$ (positive root).

$x = 1$

$f^{-1}(5) = 1$. Verification: $f(1) = 1 + 4 = 5$. Correct.

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

Solution

$y = 2(3-x)^2 + 1 = 2[-(x-3)]^2 + 1 = 2(x-3)^2 + 1$.

1. Translate right by $3$ units: $y = (x-3)^2$.
2. Vertical stretch by factor $2$: $y = 2(x-3)^2$.
3. Translate up by $1$ unit: $y = 2(x-3)^2 + 1$.

The vertex moves from $(0, 0)$ to $(3, 1)$. The parabola opens upward in both cases.

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.

Solution

First, $(g \circ f)(x) = g(f(x)) = \left(\dfrac{1}{x+1}\right)^2 = \dfrac{1}{(x+1)^2}$.

Then:

$(f \circ g \circ f)(x) = f\!\left(\frac{1}{(x+1)^2}\right) = \frac{1}{\dfrac{1}{(x+1)^2} + 1} = \frac{(x+1)^2}{(x+1)^2 + 1} = \frac{(x+1)^2}{x^2 + 2x + 2}$

Domain: need $x + 1 \neq 0 \implies x \neq -1$, and $\dfrac{1}{(x+1)^2} + 1 \neq 0$.

Since $\dfrac{1}{(x+1)^2} \geq 0$ for all $x \neq -1$, the second expression is always at least $1 > 0$.

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

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

Solution

Critical points: $2x - 1 = 0 \implies x = \dfrac{1}{2}$, and $x + 3 = 0 \implies x = -3$.

For $x \lt -3$: $f(x) = -(2x-1) + -(x+3) = -3x - 2$ (decreasing as $x$ increases).

For $-3 \leq x \lt \dfrac{1}{2}$: $f(x) = -(2x-1) + (x+3) = -x + 4$ (decreasing).

For $x \geq \dfrac{1}{2}$: $f(x) = (2x-1) + (x+3) = 3x + 2$ (increasing).

The minimum occurs at the transition from decreasing to increasing, i.e., at $x = \dfrac{1}{2}$:

$f\!\left(\frac{1}{2}\right) = 3\!\left(\frac{1}{2}\right) + 2 = \frac{7}{2}$

Minimum value: $\dfrac{7}{2}$, attained at $x = \dfrac{1}{2}$.

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

Solution

Let $y = \dfrac{x}{x^2 + 1}$. Then $yx^2 + y = x \implies yx^2 - x + y = 0$.

For real $x$, this quadratic in $x$ must have $\Delta \geq 0$:

$\Delta = 1 - 4y^2 \geq 0 \implies y^2 \leq \frac{1}{4} \implies -\frac{1}{2} \leq y \leq \frac{1}{2}$

When $y = \dfrac{1}{2}$: $\dfrac{1}{2}x^2 - x + \dfrac{1}{2} = 0 \implies (x-1)^2 = 0 \implies x = 1$. Attainable.

When $y = -\dfrac{1}{2}$: $\dfrac{1}{2}x^2 + x + \dfrac{1}{2} = 0 \implies (x+1)^2 = 0 \implies x = -1$. Attainable.

Range: $\left[-\dfrac{1}{2},\; \dfrac{1}{2}\right]$.

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 $x$.

Solution

$(g \circ f)(x) = g(2x - 1) = 2x - 1 + 3 = 2x + 2$.

$(f \circ h)(x) = f(h(x)) = 2h(x) - 1$.

Setting equal: $2h(x) - 1 = 2x + 2 \implies h(x) = x + \dfrac{3}{2}$.

Verification: $(f \circ h)(x) = 2\!\left(x + \dfrac{3}{2}\right) - 1 = 2x + 2$. Correct.

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Cross-References

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

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tip

Diagnostic Test Ready to test your understanding of Functions (Advanced)? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Functions (Advanced) with other DSE mathematics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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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, clearly 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).

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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 $\mathrm{dom}(g \circ f)$.

Solution

$\mathrm{dom}(f) = (2, \infty)$ (need $x - 2 > 0$). $\mathrm{dom}(g) = \mathbb{'\{'}R{'\}'}$.

$f \circ g$: Need $g(x) \in \mathrm{dom}(f)$, i.e., $x^2 + 1 > 2 \implies x^2 > 1 \implies x < -1$ or $x > 1$.

$\mathrm{dom}(f \circ g) = (-\infty, -1) \cup (1, \infty)$.

$g \circ f$: Need $x \in \mathrm{dom}(f) = (2, \infty)$.

$\mathrm{dom}(g \circ f) = (2, \infty)$.

Worked Example 14: Inverse of a restricted rational function

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

Solution

First, check one-to-one: $f(x) = 2 + \dfrac{6}{x - 3}$. For $x > 3$, $x - 3 > 0$, so $\dfrac{6}{x-3} > 0$ and is strictly decreasing. Therefore $f$ is strictly decreasing and hence one-to-one on $(3, \infty)$.

Set $y = \dfrac{2x}{x - 3}$:

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

$x = \frac{3y}{y - 2}$

$f^{-1}(x) = \frac{3x}{x - 2}$

To find the domain of $f^{-1}$: since $\mathrm{range}(f)$ must equal $\mathrm{dom}(f^{-1})$.

As $x \to 3^+$, $f(x) \to +\infty$. As $x \to +\infty$, $f(x) \to 2^+$.

$\mathrm{range}(f) = (2, \infty)$, so $\mathrm{dom}(f^{-1}) = (2, \infty)$.

Worked Example 15: Even and odd functions

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

Solution

$f(-x) = \frac{-x}{(-x)^2 + 1} = \frac{-x}{x^2 + 1} = -f(x)$

Since $f(-x) = -f(x)$ for all real $x$, $f$ is an odd function.

Worked Example 16: Range using discriminant

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

Solution

Let $y = \dfrac{x^2 - x + 1}{x^2 + x + 1}$. Since $x^2 + x + 1 = \left(x + \dfrac{1}{2}\right)^2 + \dfrac{3}{4} > 0$ for all $x$:

$y(x^2 + x + 1) = x^2 - x + 1 \implies (y - 1)x^2 + (y + 1)x + (y - 1) = 0$

For real $x$, $\Delta \geq 0$:

$(y + 1)^2 - 4(y - 1)^2 \geq 0 \implies (y + 1 - 2y + 2)(y + 1 + 2y - 2) \geq 0$

$(-y + 3)(3y - 1) \geq 0 \implies (y - 3)(3y - 1) \leq 0 \implies \frac{1}{3} \leq y \leq 3$

Range: $\left[\dfrac{1}{3},\; 3\right]$.

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DSE Exam-Style Questions

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

Solution

$f(2) = \dfrac{6 - 1}{2 + 2} = \dfrac{5}{4}$.

$(g \circ f)(2) = g\!\left(\dfrac{5}{4}\right) = \dfrac{5/4 + 1}{5/4 - 1} = \dfrac{9/4}{1/4} = 9$.

DSE Practice 2. The function $f$ is defined by $f(x) = 2x^2 - 8x + 5$ for $x \geq 2$. Find the range of $f$ and the value of $x$ for which $f(x) = 1$.

Solution

$f(x) = 2(x - 2)^2 - 3$. Since $x \geq 2$ and the vertex is at $x = 2$: range is $[-3, \infty)$.

$2(x - 2)^2 - 3 = 1 \implies (x - 2)^2 = 2 \implies x = 2 + \sqrt{2}$ (positive root since $x \geq 2$).

DSE Practice 3. The graph of $y = f(x)$ passes through $(0, -1)$ and has a local maximum at $(3, 4)$. Find the coordinates of the corresponding points on $y = -f(2x + 1) + 3$.

Solution

For $(0, -1)$: $2x + 1 = 0 \implies x = -\dfrac{1}{2}$. $y = -(-1) + 3 = 4$. Point: $\left(-\dfrac{1}{2},\; 4\right)$.

For the maximum at $(3, 4)$: $2x + 1 = 3 \implies x = 1$. $y = -(4) + 3 = -1$. The maximum becomes a minimum at $\left(1,\; -1\right)$.

DSE Practice 4. Let $f(x) = x^3 - 3x$. Show that $f$ is not one-to-one on $\mathbb{'\{'}R{'\}'}$, but is one-to-one on $[1, \infty)$. Find $f^{-1}(0)$.

Solution

$f(0) = 0$, $f(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0$, $f(-\sqrt{3}) = 0$. Since $f$ takes the same value at three different points, it is not one-to-one on $\mathbb{'\{'}R{'\}'}$.

For $x \geq 1$: $f'(x) = 3x^2 - 3 = 3(x^2 - 1) \geq 0$ (with equality only at $x = 1$). So $f$ is strictly increasing on $[1, \infty)$ and hence one-to-one.

$f^{-1}(0)$: solve $x^3 - 3x = 0 \implies x(x^2 - 3) = 0$. On $[1, \infty)$: $x = \sqrt{3}$. So $f^{-1}(0) = \sqrt{3}$.

DSE Practice 5. Let $h(x) = f(x) + g(x)$ where $f(x) = x^2$ and $g(x) = \dfrac{1}{x - 1}$. Find the domain of $h$.

Solution

$\mathrm{dom}(f) = \mathbb{'\{'}R{'\}'}$, $\mathrm{dom}(g) = \mathbb{'\{'}R{'\}'} \setminus \{1\}$.

$\mathrm{dom}(h) = \mathrm{dom}(f) \cap \mathrm{dom}(g) = \mathbb{'\{'}R{'\}'} \setminus \{1\}$.

DSE Practice 6. Given $f(x) = \dfrac{x}{x + 1}$ for $x \neq -1$, find $f^{-1}$ and solve $f(x) = f^{-1}(x)$.

Solution

$y = \dfrac{x}{x + 1} \implies y(x + 1) = x \implies yx + y = x \implies x(1 - y) = -y \implies x = \dfrac{y}{y - 1}$.

$f^{-1}(x) = \dfrac{x}{x - 1}$.

$f(x) = f^{-1}(x)$: $\dfrac{x}{x + 1} = \dfrac{x}{x - 1}$.

If $x = 0$: both sides equal $0$. So $x = 0$ is a solution.

If $x \neq 0$: $\dfrac{1}{x + 1} = \dfrac{1}{x - 1} \implies x - 1 = x + 1 \implies -1 = 1$, contradiction.

Solution: $x = 0$.