Functions

------- | ---------- | ------------------------------------------------ | | AQA | Paper 1, 2 | Functions, composition,” date: 2025-06-02T16:25:28.480Z tags:

• Maths
• ALevel categories:
• Maths

Definition. A function $f$ from a set $A$ (the domain) to a set $B$ (the codomain) is a Rule that assigns to each element $a \in A$ exactly one element $f(a) \in B$. We write $f: A \to B$.

Definition. The range (or image) of $f$ is the set $\{f(a) : a \in A\} \subseteq B$, the Set of all values actually attained.

Definition. The natural domain of a real-valued function defined by an algebraic expression is The largest subset of $\mathbb{R}$ for which the expression is defined. Common restrictions:

• Denominators cannot be zero: $x \neq 0$ in $\frac{1}{x}$.
• Square roots require non-negative arguments: $x \geq 0$ in $\sqrt{x}$.
• Logarithms require positive arguments: $x > 0$ in $\ln x$.

Definition. Given functions $f: A \to B$ and $g: B \to C$ The composition $g \circ f: A \to C$ is defined by:

Theorem. Function composition is associative: $(h \circ g) \circ f = h \circ (g \circ f)$.

Theorem. Function composition is not commutative : $f \circ g \neq g \circ f$.

Caution: Warning Right to left: $(f \circ g)(x) = f(g(x))$.

Definition. A function $f: A \to B$ is injective (one-to-one) if $f(a_1) = f(a_2) \implies a_1 = a_2$ for all $a_1, a_2 \in A$. Equivalently, distinct inputs give Distinct outputs.

Definition. A function $f: A \to B$ is surjective (onto) if for every $b \in B$ There exists $a \in A$ such that $f(a) = b$.

Definition. A function that is both injective and surjective is bijective.

Theorem. A function $f: A \to B$ has an inverse function $f^{-1}: B \to A$ if and only if $f$ is Bijective.

Proof.

($\Rightarrow$) If $f^{-1}$ exists, then $f$ must be injective (otherwise $f^{-1}$ would be Multiply-defined) and surjective (otherwise $f^{-1}$ would be undefined for elements not in the Range).

($\Leftarrow$) If $f$ is bijective, then for each $b \in B$ there exists exactly one $a \in A$ with $f(a) = b$. Define $f^{-1}(b) = a$. This is well-defined and satisfies $f^{-1}(f(a)) = a$ and $f(f^{-1}(b)) = b$. $\blacksquare$

Intuition. An inverse function “undoes” the original function. For this to work, the original Function must pair each input with a unique output (injectivity) and must cover every element of the Codomain (surjectivity).

Method.

1. Write $y = f(x)$.
2. Solve for $x$ in terms of $y$.
3. Replace $y$ with $x$ (and $x$ with $y$) to get $f^{-1}(x)$.
4. State the domain of $f^{-1}$ (which equals the range of $f$).

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

Proof. If $(a, b)$ lies on $y = f(x)$ Then $b = f(a)$ So $a = f^{-1}(b)$ Meaning $(b, a)$ lies On $y = f^{-1}(x)$. Swapping coordinates is reflection in $y = x$. $\blacksquare$

Definition. The modulus (absolute value) function is defined by:

Properties:

Theorem (Triangle Inequality). $|a + b| \leq |a| + |b|$ for all $a, b \in \mathbb{R}$.

Proof. We consider cases based on the signs of $a$ and $b$.

Case 1: $a, b \geq 0$. Then $|a + b| = a + b = |a| + |b|$. Equality holds.

Case 2: $a, b \leq 0$. Then $a + b \leq 0$ So $|a + b| = -(a + b) = -a - b = |a| + |b|$. Equality Holds.

Case 3: $a \geq 0$, $b \leq 0$. If $a + b \geq 0$: $|a + b| = a + b \leq a + (-b) = |a| + |b|$. If $a + b \leq 0$: $|a + b| = -(a + b) = -a - b \leq a - b = |a| + |b|$ (since $a \geq 0$ implies $a \geq -a$ I.e., $2a \geq 0$). $\blacksquare$

To solve $|f(x)| = g(x)$:

1. Case 1: $f(x) \geq 0$ So $f(x) = g(x)$.
2. Case 2: $f(x) < 0$ So $-f(x) = g(x)$.
3. Check solutions against the case conditions.

Theorem. The graph of $y = f(x - a)$ is the graph of $y = f(x)$ translated by $a$ units in the Positive $x$-direction.

Proof. Let $g(x) = f(x - a)$. The point $(x_0, y_0)$ lies on $y = g(x)$ if and only if $y_0 = g(x_0) = f(x_0 - a)$. This means $(x_0 - a, y_0)$ lies on $y = f(x)$. So the point $(x_0, y_0)$ on $g$ corresponds to the point $(x_0 - a, y_0)$ on $f$, a shift right by $a$. $\blacksquare$

Intuition. Replacing $x$ with $x - a$ means “to get the same output, I need to input $a$ more.” The graph shifts right to compensate.

Similarly, $y = f(x) + b$ translates up by $b$.

Proof for $y = f(-x)$ as reflection in the $y$-axis. If $(x_0, y_0)$ is on $y = f(x)$ Then $y_0 = f(x_0)$. The point $(-x_0, y_0)$ satisfies $y_0 = f(-(-x_0)) = f(x_0)$ So it lies on $y = f(-x)$. Reflecting $(x_0, y_0)$ in the $y$-axis gives $(-x_0, y_0)$. $\blacksquare$

Proof for $y = f(ax)$ as horizontal stretch. If $(x_0, y_0)$ is on $y = f(x)$ Then on $y = f(ax)$ The same $y$-value occurs when $ax = x_0$ I.e., $x = \frac{x_0}{a}$. So $(\frac{x_0}{a}, y_0)$ is on the new graph, a horizontal stretch by factor $\frac{1}{a}$. $\blacksquare$

When multiple transformations are applied, the order matters. The convention is:

Represents: horizontal translation by $p$ (right), vertical stretch by factor $a$ Vertical Translation by $q$ (up).

Tip: Tip (outside). The order inside-out matters.

Definition. A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain. A function $f$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.

Theorem. The graph of an even function is symmetric about the $y$-axis. The graph of an odd Function has rotational symmetry of order 2 about the origin.

Proof. For even $f$: the point $(-x, f(-x)) = (-x, f(x))$ is on the graph whenever $(x, f(x))$ is. These two points are reflections of each other across the $y$-axis. $\blacksquare$

For odd $f$: the point $(-x, f(-x)) = (-x, -f(x))$ is on the graph whenever $(x, f(x))$ is. Rotating $(x, f(x))$ by $180^\circ$ about the origin gives $(-x, -f(x))$. $\blacksquare$

Algebraic Properties.

Proof (for $f$ odd, $g$ odd $\implies fg$ even). $(fg)(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) = (fg)(x)$. $\blacksquare$

Proof (for $f$ even, $g$ odd $\implies fg$ odd). $(fg)(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -(fg)(x)$. $\blacksquare$

Theorem. The only function that is both even and odd is $f(x) = 0$ (the zero function on a Symmetric domain).

Proof. $f$ even: $f(-x) = f(x)$. $f$ odd: $f(-x) = -f(x)$. Therefore $f(x) = -f(x)$ So $2f(x) = 0$ Hence $f(x) = 0$ for all $x$. $\blacksquare$

Theorem. The domain of $g \circ f$ is $\{x \in \mathrm{dom}(f) : f(x) \in \mathrm{dom}(g)\}$.

Proof. For $(g \circ f)(x) = g(f(x))$ to be defined, we need $x \in \mathrm{dom}(f)$ (so $f(x)$ Exists) AND $f(x) \in \mathrm{dom}(g)$ (so $g$ can accept $f(x)$ as input). $\blacksquare$

The range of $g \circ f$ is the image under $g$ of the set $\{f(x) : x \in \mathrm{dom}(g \circ f)\}$ Which is a subset of the range of $g$.

Definition. A function $f$ is self-inverse if $f^{-1} = f$ I.e., $f(f(x)) = x$ for all $x$ In the domain of $f$.

Theorem. If $f$ is self-inverse, then $f$ is bijective and $f = f^{-1}$.

Proof. If $f(f(x)) = x$ for all $x$ Then $f$ has an inverse (namely $f$ itself), so $f$ is Bijective. And by definition of inverse, $f^{-1} = f$. $\blacksquare$

Common self-inverse functions:

• $f(x) = \frac{a}{x}$ (reciprocal, $x \neq 0$): $f(f(x)) = a/(a/x) = x$. ✓
• $f(x) = a - x$ (reflection): $f(f(x)) = a - (a - x) = x$. ✓
• $f(x) = \frac{ax + b}{cx - a}$ (general fractional linear, $c \neq 0$): verify $f(f(x)) = x$.

Proof for $f(x) = (ax+b)/(cx-a)$. Let $f(x) = \frac{ax+b}{cx-a}$. Then:

$\blacksquare$ (provided $a^2 + bc \neq 0$).

To solve inequalities involving the modulus function:

Type 1: $|f(x)| \lt g(x)$ where $g(x) > 0$.

This is equivalent to the compound inequality $-g(x) \lt f(x) \lt g(x)$.

Proof. $|f(x)| \lt g(x) \iff -g(x) \lt f(x) \lt g(x)$ follows directly from the definition of Modulus. $\blacksquare$

Type 2: $|f(x)| \gt g(x)$.

This is equivalent to: $f(x) > g(x)$ OR $f(x) < -g(x)$.

When multiple transformations are applied to $y = f(x)$ The order matters because horizontal and Vertical transformations interact differently.

Rule. For $y = af(bx + c) + d$:

1. Apply horizontal transformations first: the argument is $bx + c$ Which is a horizontal stretch by factor $1/b$ then a horizontal shift of $-c/b$.
2. Apply vertical transformations second: vertical stretch by $|a|$ Reflection if $a < 0$ Then vertical shift $d$.

Problem 1. Given $f(x) = \frac{1}{x - 3} + 2$ Find the domain, range, and inverse function.

Problem 2. Given $f(x) = x^2 - 4x + 9$ for $x \geq 2$ Find $f^{-1}(x)$.

Problem 3. Solve $|3x + 1| = 2x + 5$.

Problem 4. Given $f(x) = 2x - 1$ and $g(x) = x^2 + 3$ Find $(g \circ f)(x)$ and solve $(g \circ f)(x) = 12$.

Problem 5. The graph of $y = f(x)$ passes through $(0, 1)$ and $(3, -2)$. State the coordinates Of the corresponding points on: (a) $y = f(x + 2)$ (b) $y = -f(x)$ (c) $y = f(2x)$ (d) $y = 3f(x) - 1$

Problem 6. The function $f$ is defined by $f(x) = x^3 - 3x + 1$. Show that $f$ is not injective On $\mathbb{R}$ And find the largest interval containing $x = 0$ on which $f$ is injective.

Problem 7. Solve the inequality $|x - 3| > |2x + 1|$.

Problem 8. Given $f(x) = e^{2x}$ and $g(x) = \ln(x + 1)$ Find f^{-1}$$g^{-1} And the Domain of $f \circ g$.

Problem 9. Sketch the graph of $y = |x^2 - 4|$ Showing the coordinates of all points where the Graph meets the axes.

Problem 10. Prove that $f: [0, \infty) \to [0, \infty)$ defined by $f(x) = x^2 + 4x$ is Bijective, and find $f^{-1}$.

Problem 11. Classify each function as even, odd, or neither: (a) $f(x) = x^4 - x^2$ (b) $g(x) = x^3 + x$ (c) $h(x) = x + 1$ (d) $k(x) = |x|$

Problem 12. Given $f(x) = 2x + 3$ with domain $\mathbb{R}$ and $g(x) = \sqrt{x - 1}$ with domain $[1, \infty)$ Find the domain of $f \circ g$ and $g \circ f$.

Problem 13. Verify that $f(x) = \frac{3x + 2}{x - 3}$ is self-inverse.

Problem 14. Solve the inequality $|x^2 - 5x + 6| \geq |x - 2|$.

Problem 15. The graph of $y = f(x)$ passes through $(1, 3)$ and $(-2, 5)$. State the coordinates Of the corresponding points on the graph of $y = 2f(3x - 1) + 4$.

• Differentiation — Derivatives describe the rate of change of functions and are used to find stationary points and sketch graphs.
• Exponentials and Logarithms — The exponential and logarithmic functions are inverses of each other, a key example of function inversion.
• Integration — Integration finds areas under function curves and reverses the process of differentiation.
• Sequences and Series — Sequences are functions defined on the natural numbers, linking function behaviour to series convergence.

1. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

2. Losing marks by not showing sufficient working. Always write out each step, especially in proof questions.

3. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

4. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

flowchart TD
    A[05 Functions] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.