UK A-Level: Functions and Graphs

Mastering functions and their graphical representations is the cornerstone of advanced mathematics, providing the language to model everything from simple motion to complex economic behaviour. This topic is not just a collection of abstract rules; it is a powerful toolkit for understanding how quantities relate to each other, forming the foundation for calculus, mechanics, and further pure mathematics you will encounter.

Function Fundamentals: Domain, Range, and Notation

At its core, a function is a rule that assigns each input from a specific set to exactly one output. We use notation like $f(x)=x^2+3$ or $g:x\mapsto \frac{1}{x-2}$. The set of all possible input values is called the domain. If not stated, the domain is the largest set of real numbers for which the function is defined. For $f(x)=\sqrt{x-1}$, the expression under the square root must be non-negative, so $x-1\ge 0$, giving a domain of $x\ge 1$ or $[1,\infty )$.

The range is the set of all possible output values, determined by applying the function to every element in the domain. For $f(x)=x^2+3$ with a domain of all real numbers, $x^2$ is always $\ge 0$, so the smallest output is $3$. The range is therefore $f(x)\ge 3$ or $[3,\infty )$. Understanding domain and range from the outset prevents impossible operations and is crucial for working with composite and inverse functions later.

Composite Functions and Finding Inverses

We can combine functions. The composite function $fg(x)$ (or $f\circ g(x)$) means you apply $g$ first, then apply $f$ to the result: $fg(x)=f(g(x))$. The order matters profoundly: $fg(x)$ is rarely the same as $gf(x)$. To construct it, you substitute the entire expression for $g(x)$ into the input of $f(x)$. If $f(x)=2x+1$ and $g(x)=x^2$, then $fg(x)=f(g(x))=f(x^2)=2(x^2)+1=2x^2+1$.

An inverse function, denoted $f^{-1}(x)$, reverses the action of $f$. If $f(a)=b$, then $f^{-1}(b)=a$. A function must be one-to-one (each $y$ value comes from only one $x$ value) to have a full inverse. Graphically, $f^{-1}(x)$ is a reflection of the graph of $f(x)$ in the line $y=x$.

To find the inverse analytically, follow these steps:

1. Write the function as $y=f(x)$.
2. Swap $x$ and $y$ to get $x=f(y)$.
3. Rearrange this new equation to make $y$ the subject.
4. Express the inverse as $f^{-1}(x)=$ (the new expression in $x$).

For $f(x)=\frac{2x+3}{5}$, we write $y=\frac{2x+3}{5}$, swap to get $x=\frac{2y+3}{5}$, and solve for $y$: $5x=2y+3$, so $2y=5x-3$, hence $y=\frac{5x-3}{2}$. Therefore, $f^{-1}(x)=\frac{5x-3}{2}$.

The Modulus Function and Related Equations

The modulus function, written $|x|$, outputs the absolute value of $x$. It is defined as: $$|x|=\{\begin{array}{ll}x,& \text{if }x\ge 0\\ -x,& \text{if }x<0\end{array}$$

Graphically, $y=|x|$ is a V-shaped graph with its vertex at the origin. The graph of $y=|f(x)|$ takes the graph of $y=f(x)$ and reflects any part below the $x$-axis (where $f(x)$ is negative) upwards into the positive $y$ region. Conversely, $y=f(|x|)$ reflects the part of the graph for $x\ge 0$ into the $x<0$ region, creating a symmetrical graph about the $y$-axis.

Solving modulus equations often requires considering separate cases to remove the modulus signs. To solve $|2x-1|=5$, you consider the two possibilities defined by the function: Case 1: $2x-1\ge 0$, so $|2x-1|=2x-1$. Equation becomes $2x-1=5\Rightarrow x=3$. Case 2: $2x-1<0$, so $|2x-1|=-(2x-1)=-2x+1$. Equation becomes $-2x+1=5\Rightarrow x=-2$. Always check your solutions satisfy the initial case condition. Here, $3\ge 0$ and $-2<0$, so both are valid: $x=3$ or $x=-2$.

Graph Transformations

Transformations allow you to sketch complex functions by modifying a known basic graph. The key transformations are defined relative to a base function $y=f(x)$.

1. Translations (Shifts):

• $y=f(x)+a$ is a vertical translation by $+a$ units. Every point moves up (for $a>0$) or down.
• $y=f(x+a)$ is a horizontal translation by $-a$ units. The graph moves left for $a>0$ and right for $a<0$. This is often the most counterintuitive rule.

1. Stretches:

• $y=af(x)$ is a vertical stretch by scale factor $a$. The $y$-coordinates are multiplied by $a$.
• $y=f(ax)$ is a horizontal stretch by scale factor $\frac{1}{a}$. The graph is squashed towards the $y$-axis if $a>1$.

1. Reflections:

• $y=-f(x)$ reflects the graph in the $x$-axis.
• $y=f(-x)$ reflects the graph in the $y$-axis.

Combining Multiple Transformations

To apply multiple transformations, you must follow a consistent order, typically working from the inside of the function notation outwards. Consider transforming the graph of $y=\sqrt{x}$ to sketch $y=3-2\sqrt{x+1}$.

First, rewrite the equation in the standard transformed form: $y=-2\sqrt{x+1}+3$. Start with the basic graph $y=\sqrt{x}$.

1. Inside the function: $y=\sqrt{x+1}$. This is a horizontal translation by $-1$ (1 unit to the left).
2. Next, apply the vertical stretch: $y=2\sqrt{x+1}$. Scale factor 2, parallel to the $y$-axis.
3. Apply the reflection: $y=-2\sqrt{x+1}$. This reflects the stretched graph in the $x$-axis.
4. Finally, apply the vertical translation: $y=-2\sqrt{x+1}+3$. Shift the entire graph up by 3 units.

The key is to perform the transformations that affect the $x$-variable (inside $f$) first, in the order they are applied, then do those affecting the $y$-variable (outside $f$).

Common Pitfalls

1. Misordering Composite Functions: Confusing $fg(x)$ with $gf(x)$ is a frequent error. Remember, the notation $fg(x)$ means apply the function closest to the $x$ first. Correction: Always read it as $f(g(x))$. Write the inner function $g(x)$ first, then substitute it into the outer function $f$.

1. Incorrect Horizontal Transformations: Thinking $f(x+2)$ shifts the graph to the right is a classic mistake. Correction: Associate the sign inside the bracket with the opposite direction. $f(x+2)$ shifts left by 2. For stretches, $f(2x)$ is a squash by factor 1/2, not 2.

1. Swapping Incorrectly When Finding Inverses: Some students try to rearrange first and then swap variables, leading to algebraic chaos. Correction: Always follow the process methodically: 1) $y=f(x)$, 2) swap $x$ and $y$, 3) rearrange to $y=...$.

1. Forgetting to Check Solutions for Modulus Equations: After solving the cases for an equation like $|3x-4|=x$, you might get candidate solutions that do not satisfy the initial conditions of the case you solved. Correction: For each solution, verify it works in the original equation or at least meets the condition (e.g., $3x-4\ge 0$) of the case you used to find it.

Summary

• The domain is the set of allowed inputs, and the range is the resulting set of outputs. Determining these is essential for understanding a function's behaviour.
• Composite functions $fg(x)$ apply one function after another; order is critical. The inverse function $f^{-1}(x)$ reverses the process and is found by swapping variables and rearranging.
• The modulus function $|x|$ outputs the absolute value. Its graph is V-shaped, and solving modulus equations requires considering separate positive and negative cases.
• Graph transformations follow specific rules: $f(x)+a$ (vertical shift), $f(x+a)$ (horizontal shift), $af(x)$ (vertical stretch), $f(ax)$ (horizontal stretch), and reflections with negative signs.
• When combining transformations, a systematic inside-out approach is necessary. Rewrite the equation, then apply changes to the $x$-variable (inside the function) before changes to the $y$-variable (outside).