Pure Mathematics 1 (Paper 1) Study Notes: Functions (1.2)
Welcome to the world of Functions! This chapter is absolutely vital—it forms the backbone for topics like differentiation, integration, and advanced maths in general. A solid understanding here will make your journey through AS/A Level much smoother. Let’s dive in!
1. Defining a Function, Domain, and Range
A function is simply a rule that links an input to an output, but with one crucial condition: for every single input, there must be exactly one output.
Analogy: Think of a function \(f\) as a high-tech vending machine. If you press the button for input 'A', you only ever get output 'B'. If pressing 'A' sometimes gave 'B' and sometimes 'C', it wouldn't be a function!
Key Terms to Know
- Function Notation: We write \(y = f(x)\). For example, \(f(x) = x^2 - 1\) or \(f: x \to x^2 - 1\).
- Domain: This is the set of all possible input values (\(x\)-values) that the function can accept.
- Range: This is the set of all actual output values (\(y\)-values) produced by the function using its domain.
How to Determine the Domain (Inputs)
Generally, the domain is assumed to be all real numbers (\(x \in \mathbb{R}\)) unless two common problems exist:
- Division by zero: The denominator of any fraction cannot be zero.
Example: For \(f(x) = \frac{1}{x-3}\), the domain cannot include \(x=3\). - Square roots of negative numbers: You cannot take the square root of a negative number (in Pure 1).
Example: For \(g(x) = \sqrt{x+5}\), the domain must be \(x \ge -5\).
How to Determine the Range (Outputs)
The range is often easier to find by sketching the graph or using algebraic techniques like completing the square (especially for quadratic functions).
Example: Given \(f(x) = x^2 + 4\) where the domain is \(x \in \mathbb{R}\).
- Since \(x^2 \ge 0\), the smallest possible value of \(x^2 + 4\) is \(0 + 4 = 4\).
- The range is therefore \(f(x) \ge 4\).
Always define the domain first. The range is the set of outputs generated from that specific domain.
2. One-One and Many-One Functions
The type of function determines if we can find an inverse. We primarily focus on two types:
One-One Function
A function is one-one (or injective) if every distinct input maps to a distinct output. No two inputs ever share the same output value.
- Example: \(f(x) = 2x + 1\)
- Test: Use the Horizontal Line Test. If any horizontal line crosses the graph more than once, it is NOT one-one.
Many-One Function
A function is many-one (or surjective) if two or more inputs can map to the same output.
- Example: \(f(x) = x^2\). Both \(x=2\) and \(x=-2\) give \(f(x)=4\).
Only one-one functions can have an inverse function. If a function is many-one (like \(y=x^2\)), you must restrict its domain to make it one-one before finding the inverse.
3. Composite Functions
A composite function is formed when you apply one function after another. It's the process of putting the output of one function directly into the input of a second function.
Notation and Order
- If you have functions \(f\) and \(g\):
- \(fg(x)\): This means apply \(g\) first, then apply \(f\) to the result. (Work from the inside out: \(f(g(x))\)).
- \(gf(x)\): This means apply \(f\) first, then apply \(g\) to the result. (\(g(f(x))\)).
- \(f^2(x)\) or \(ff(x)\): This means apply \(f\) to the result of \(f(x)\).
Step-by-Step Example
Let \(f(x) = 3x - 1\) and \(g(x) = x^2\). Find \(fg(x)\).
- Start with \(f(g(x))\). Replace \(g(x)\) with its definition: \(f(x^2)\).
- Now, look at function \(f\). \(f\) takes its input and multiplies it by 3, then subtracts 1.
- Since the input is now \(x^2\), we get: \(3(x^2) - 1\).
- Result: \(fg(x) = 3x^2 - 1\).
The Compatibility Constraint (Syllabus Key Point)
For the composite function \(fg\) to be successfully formed, the range of \(g\) must be contained within the domain of \(f\).
Did you know? If \(g\) outputs the number 10, but \(f\) is defined only for inputs less than 5, then the composition \(fg\) cannot exist (or its domain must be restricted).
Order matters! \(fg(x)\) is almost never the same as \(gf(x)\). Work from the inside out.
4. Inverse Functions (\(f^{-1}\))
The inverse function reverses the process of \(f(x)\). If \(f\) takes 2 to 5, then \(f^{-1}\) takes 5 back to 2.
The Importance of Domain and Range Swaps
- The Domain of \(f^{-1}\) is the Range of \(f\).
- The Range of \(f^{-1}\) is the Domain of \(f\).
This is crucial for defining the inverse correctly, especially when dealing with restricted domains.
Step-by-Step: Finding the Inverse Algebraically
Find the inverse of \(h(x) = (2x + 3)^2 - 4\) for the domain \(x < -1.5\).
- Set \(y = h(x)\): \(y = (2x + 3)^2 - 4\).
- Swap \(x\) and \(y\): \(x = (2y + 3)^2 - 4\).
- Isolate \(y\) (Rearrange):
- \(x + 4 = (2y + 3)^2\)
- \(\pm \sqrt{x+4} = 2y + 3\)
- Determine the Sign and Final Range:
- The original domain was \(x < -1.5\). This means \(2x + 3 < 0\).
- Since \(2y+3\) must be negative (as \(y\) is the output/range of the inverse, which must match the original domain), we choose the negative square root.
- \(- \sqrt{x+4} = 2y + 3\)
- \(2y = -3 - \sqrt{x+4}\)
- \(y = \frac{-3 - \sqrt{x+4}}{2}\)
- State the Inverse: \(h^{-1}(x) = \frac{-3 - \sqrt{x+4}}{2}\).
Always check the original domain when solving for the square root in step 3. That restriction determines which sign (\(+\) or \(-\)) you must choose for the inverse!
Graphical Relationship
The graph of \(y = f^{-1}(x)\) is a reflection of the graph of \(y = f(x)\) in the line \(y = x\).
Visualize: Fold your paper along the line \(y=x\). The two graphs should match up perfectly!
5. Function Transformations
Transformations describe how moving or stretching a basic graph \(y = f(x)\) creates a new graph.
Don't worry if this seems tricky at first—just remember that transformations happening inside the bracket are usually counter-intuitive (the opposite of what you expect)!
Memory Aid: Inside vs. Outside
- Outside the bracket/function (\(f(x) + a\)): Affects the \(y\)-values. It’s Vertical, and it works intuitively (plus means up, minus means down).
- Inside the bracket/function (\(f(x + a)\)): Affects the \(x\)-values. It’s Horizontal, and it works non-intuitively (plus means left, minus means right).
| Transformation | Effect on Graph | Type of Action | Terminology |
| \(y = f(x) + a\) | Vertical shift by vector \(\begin{pmatrix} 0 \\ a \end{pmatrix}\) | Translation | Move up if \(a>0\), down if \(a<0\). |
| \(y = f(x + a)\) | Horizontal shift by vector \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\) | Translation | Move left if \(a>0\), right if \(a<0\). |
| \(y = af(x)\) | Multiplies all \(y\)-coordinates by \(a\) | Stretch (Vertical) | Stretch parallel to the \(y\)-axis by factor \(a\). If \(a=-1\), it’s a reflection in the x-axis. |
| \(y = f(ax)\) | Divides all \(x\)-coordinates by \(a\) | Stretch (Horizontal) | Stretch parallel to the \(x\)-axis by factor \(1/a\). If \(a=-1\), it’s a reflection in the y-axis. |
Simple Combinations
When multiple transformations occur, the order is important, particularly when both horizontal and vertical stretches/translations are involved. However, for simple combinations, follow the order: Stretches/Reflections first, then Translations.
Example: Transforming \(y = f(x)\) to \(y = 2f(x) - 3\).
- Vertical Stretch by factor 2 (from \(2f(x)\)).
- Vertical Translation by vector \(\begin{pmatrix} 0 \\ -3 \end{pmatrix}\) (from \(-3\)).
Outside = Vertical (Intuitive). Inside = Horizontal (Counter-intuitive: factor \(1/a\)). Apply Stretches/Reflections before Translations.