Hello Future P3 Mathematicians!

Welcome to the core chapter on Algebra and Functions for Unit P3! If you thought you were done with functions after P1 and P2, think again! P3 takes these foundational concepts—like domain, range, and graphs—and introduces sophisticated tools like the Modulus Function, complex Function Combinations, and the powerful technique of Partial Fractions.

This chapter is crucial. The concepts here underpin calculus, integration, and even differential equations later on. Don't worry if some parts look tricky; we will break everything down into manageable, easy-to-understand steps. Let's make P3 functions click!


1. The Modulus Function (Absolute Value)

The modulus function, denoted by \(|x|\), gives the absolute value of a number. This means it always returns the non-negative value of the input.

1.1 Defining the Modulus

The definition is piece-wise:

  • If \(x \ge 0\), then \(|x| = x\)
  • If \(x < 0\), then \(|x| = -x\)

Analogy: Think of the modulus function as calculating distance. Whether you walk 5km North (+5) or 5km South (-5), the distance you traveled is always positive 5km. It removes the direction (sign).

Key Term: The term absolute value is interchangeable with the modulus function.

1.2 Graphing Modulus Functions

You need to understand two main types of modulus transformations:

Case A: Graphing \(y = |f(x)|\)

This affects the output (\(y\) values). Any part of the original graph \(y=f(x)\) that falls below the \(x\)-axis (where \(y\) is negative) is reflected upwards across the \(x\)-axis.

Step-by-Step for \(y = |f(x)|\):

  1. Draw the original graph \(y = f(x)\).
  2. Identify the section(s) below the \(x\)-axis.
  3. Reflect these sections in the \(x\)-axis.
Case B: Graphing \(y = f(|x|)\)

This affects the input (\(x\) values). Since \(|x|\) means you always feed a positive value into \(f\), the graph must be symmetrical about the \(y\)-axis.

Step-by-Step for \(y = f(|x|)\):

  1. Ignore the left side (negative \(x\) values) of the original graph \(y=f(x)\).
  2. Keep the right side (positive \(x\) values and \(x=0\)).
  3. Reflect the right side across the \(y\)-axis to create the left side of the new graph.

1.3 Solving Modulus Equations and Inequalities

The safest way to solve these is to consider the two separate cases (the positive input and the negative input) or, for equations, the squaring method.

Solving Equations (e.g., \(|2x - 1| = x + 4\))

Method 1: Two Cases

  • Case 1 (Positive): \(2x - 1 = x + 4 \implies x = 5\)
  • Case 2 (Negative): \(-(2x - 1) = x + 4 \implies -2x + 1 = x + 4 \implies -3 = 3x \implies x = -1\)
  • Crucial Check: Always substitute your answers back into the original equation to ensure they work, especially if \(x\) is on both sides. (Both \(x=5\) and \(x=-1\) work here.)

Method 2: Squaring Both Sides (Highly Recommended for safety)

If \(|A| = B\), then \(A^2 = B^2\). This removes the modulus signs completely.

Example: \(|2x - 1| = x + 4\)

\((2x - 1)^2 = (x + 4)^2\)
\(4x^2 - 4x + 1 = x^2 + 8x + 16\)
\(3x^2 - 12x - 15 = 0\)
\(x^2 - 4x - 5 = 0\)
\((x - 5)(x + 1) = 0\)

Solutions: \(x = 5\) and \(x = -1\).

Solving Inequalities

It is often easiest to solve inequalities by squaring both sides or by sketching the graphs.

Rule 1: Less Than (\(<\) or \(\le\))

If \(|x| < a\), then \(-a < x < a\). (The "Sandwich" inequality)

Example: \(|2x - 1| \le 5\)
\(-5 \le 2x - 1 \le 5\)
\(-4 \le 2x \le 6\)
\(-2 \le x \le 3\)

Rule 2: Greater Than (\(>\) or \(\ge\))

If \(|x| > a\), then \(x > a\) OR \(x < -a\). (The "Outer" solution)

Example: \(|3x + 2| > 4\)
Case 1: \(3x + 2 > 4 \implies 3x > 2 \implies x > 2/3\)
Case 2: \(3x + 2 < -4 \implies 3x < -6 \implies x < -2\)
Solution: \(x < -2\) or \(x > 2/3\)

Quick Review: Modulus

The modulus makes the result positive. When solving, use the two cases (\(A=B\) and \(A=-B\)) or square both sides. Always check your answers!


2. Combining Functions

In P3, we combine functions not just by adding or multiplying, but by feeding the output of one function directly into the input of another. This is called composition.

2.1 Function Notation Review: Domain and Range

  • Domain: The set of allowed input values (\(x\)).
  • Range: The set of resulting output values (\(y\)) after the function has acted on the domain.

A function \(f\) is often defined as \(f: x \mapsto f(x)\) for a specific domain. For example, \(f: x \mapsto \sqrt{x}\), \(x \in \mathbb{R}, x \ge 0\). (The domain is \(x \ge 0\)).

2.2 Composite Functions

When you have two functions, \(f(x)\) and \(g(x)\), the composite function \(fg(x)\) means you apply \(g\) first, and then apply \(f\) to the result.

Notation: \(fg(x)\) means \(f(g(x))\).
Order matters! \(fg(x)\) is generally NOT the same as \(gf(x)\).

Example: Let \(f(x) = 2x + 1\) and \(g(x) = x^2\).

\(fg(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1\)

\(gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1\)

2.3 Domain of Composite Functions

This is where students often make mistakes. For \(fg(x)\) to exist, the range of the inner function (\(g\)) must be a subset of the domain of the outer function (\(f\)).

The domain of \(fg\) is the set of inputs \(x\) for \(g\) such that the output \(g(x)\) is a valid input for \(f\).

Example requiring caution:
Let \(f(x) = \frac{1}{x}\) and \(g(x) = 3 - x\). The domain of \(f\) is \(x \ne 0\).

\(fg(x) = f(3 - x) = \frac{1}{3 - x}\).
For \(fg(x)\) to be defined, the output of \(g(x)\) cannot be zero.
\(g(x) = 0 \implies 3 - x = 0 \implies x = 3\).
Therefore, the domain of \(fg\) is \(x \ne 3\).

Key Takeaway: Composition

Composition means "nesting" one function inside another: \(f(g(x))\). Remember to check the domain restriction on the inner function's output.


3. Inverse Functions

An inverse function, \(f^{-1}(x)\), reverses the action of \(f(x)\). If \(f\) turns 5 into 10, \(f^{-1}\) turns 10 back into 5.

3.1 The Condition for an Inverse

A function can only have an inverse if it is a one-to-one function. This means every output (\(y\)) corresponds to only one input (\(x\)).

  • Test: Use the Horizontal Line Test. If any horizontal line crosses the graph of \(f(x)\) more than once, the function is NOT one-to-one, and therefore has no inverse over that domain.
  • Quadratic functions (parabolas) are usually many-to-one. To find an inverse for \(f(x) = x^2\), we must restrict the domain (e.g., to \(x \ge 0\)).

3.2 Finding the Inverse Function \(f^{-1}(x)\)

This is a mechanical, step-by-step process:

Step-by-Step Example: Find \(f^{-1}(x)\) for \(f(x) = \frac{3x - 1}{x + 2}\)

  1. Replace \(f(x)\) with \(y\): \[ y = \frac{3x - 1}{x + 2} \]
  2. Swap \(x\) and \(y\): (This is the definition of the inverse operation) \[ x = \frac{3y - 1}{y + 2} \]
  3. Rearrange to make \(y\) the subject: \[ x(y + 2) = 3y - 1 \] \[ xy + 2x = 3y - 1 \]
  4. Gather terms containing \(y\) on one side, and terms without \(y\) on the other: \[ xy - 3y = -1 - 2x \]
  5. Factor out \(y\): \[ y(x - 3) = -1 - 2x \]
  6. Divide to isolate \(y\): \[ y = \frac{-1 - 2x}{x - 3} \]
  7. Replace \(y\) with \(f^{-1}(x)\): \[ f^{-1}(x) = \frac{-1 - 2x}{x - 3} \]

Did you know? A rational function like the one above is called self-inverse if \(f(x) = f^{-1}(x)\). This happens when the function's graph is symmetric about \(y=x\).

3.3 Domains and Ranges of Inverses

The domain and range flip when you move from a function to its inverse:

  • The Domain of \(f\) is the Range of \(f^{-1}\).
  • The Range of \(f\) is the Domain of \(f^{-1}\).

Memory Trick: DR = RD (Domain/Range flips Range/Domain).

3.4 Graphs of Inverses

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

Quick Review: Inverses

Requirement: Must be one-to-one. Find it by swapping \(x\) and \(y\) and rearranging. The graph is reflected in \(y=x\).


4. Partial Fractions

Partial fractions is a technique used to decompose a single, complicated fraction (usually one involving polynomials) into a sum of two or more simpler fractions. This is vital because these simpler fractions are often easier to integrate (a skill you'll use extensively in P4).

4.1 Proper vs. Improper Fractions

Partial fraction decomposition only works directly on proper fractions.

  • Proper Fraction: The degree (highest power) of the numerator is less than the degree of the denominator. (e.g., \(\frac{x}{x^2+1}\)).
  • Improper Fraction: The degree of the numerator is greater than or equal to the degree of the denominator. (e.g., \(\frac{x^2+3}{x^2-1}\)).

If you have an improper fraction, you MUST perform algebraic long division (or inspection) first! This results in a polynomial plus a proper fraction remainder: \[ \frac{\text{Numerator}}{\text{Denominator}} = \text{Polynomial} + \frac{\text{Remainder}}{\text{Denominator}} \]

4.2 Three Main Forms of Decomposition (P3 Focus)

We assume the denominator has been fully factorized.

Form 1: Distinct Linear Factors

If the denominator contains factors like \((ax+b)(cx+d)\), the decomposition is:

\[ \frac{P(x)}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d} \]
Form 2: Repeated Linear Factors

If the denominator contains a squared factor like \((ax+b)^2\), we need two separate terms for this factor:

\[ \frac{P(x)}{(ax+b)^2(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{cx+d} \]

Note: You MUST include the single power term, \(A/(ax+b)\).

Form 3: Irreducible Quadratic Factors (Brief mention)

If the denominator contains a quadratic factor that cannot be factored further (e.g., \(x^2+4\)), the numerator above it must be linear:

\[ \frac{P(x)}{(ax^2+bx+c)(dx+e)} \equiv \frac{Ax+B}{ax^2+bx+c} + \frac{C}{dx+e} \]

4.3 The Process of Decomposition

To find the unknown constants (A, B, C, etc.), you have two main methods:

Method 1: Substitution (The Quick Way)

This method is best when you have distinct linear factors. Once you have set up the identity, choose values of \(x\) that make the denominator factors zero. This zeroes out one or more constants, allowing you to solve for the remaining ones easily.

Step-by-Step Example (Distinct Factors):
Decompose \(\frac{5x - 1}{(x - 1)(x + 2)}\).

  1. Set up the identity: \[ \frac{5x - 1}{(x - 1)(x + 2)} \equiv \frac{A}{x - 1} + \frac{B}{x + 2} \]
  2. Multiply by the common denominator: \[ 5x - 1 \equiv A(x + 2) + B(x - 1) \]
  3. Substitute \(x = 1\) (to eliminate \(B\)): \[ 5(1) - 1 = A(1 + 2) + B(0) \] \[ 4 = 3A \implies A = 4/3 \]
  4. Substitute \(x = -2\) (to eliminate \(A\)): \[ 5(-2) - 1 = A(0) + B(-2 - 1) \] \[ -11 = -3B \implies B = 11/3 \]
  5. Write the final decomposition: \[ \frac{4/3}{x - 1} + \frac{11/3}{x + 2} \]
Method 2: Comparing Coefficients

This is necessary when factors cannot be zeroed out (e.g., with irreducible quadratics) or as a backup for repeated factors.

  1. Set up the identity and expand (as in step 2 above).
  2. Expand the right-hand side fully and group terms by powers of \(x\). \[ 5x - 1 \equiv (A+B)x + (2A - B) \]
  3. Compare coefficients of matching powers of \(x\):
    • Coefficient of \(x^1\): \(5 = A + B\)
    • Coefficient of \(x^0\) (Constant): \(-1 = 2A - B\)
  4. Solve the resulting simultaneous equations for A and B. (Adding the two equations: \(4 = 3A \implies A = 4/3\). Substitute back to find \(B=11/3\)).

Tip for Struggling Students: For repeated factors (Form 2), use a combination: use substitution to find constants related to the highest power factor (like \(B\) and \(C\)), and then use comparing coefficients for the remaining constants (like \(A\)).

Key Takeaway: Partial Fractions

Partial fractions break down complex fractions. If the numerator degree \(\ge\) denominator degree, divide first. Use Substitution (quick zeros) and Comparing Coefficients (systematic) to find the unknown constants.


You have now tackled the heavy algebra and function work of P3! Keep practicing the algebraic rearrangement for inverses and partial fractions—these mechanical skills are crucial for exam speed and accuracy. Keep going!