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Mathematical Methods · Unit 3 · AOS 1

Functions, relations and graphsUnderstand it. Then apply it.

Learn to read a function as a rule, a graph and a set of allowed inputs. Work through six connected ideas, then build a graph and check your reasoning on fresh questions.

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1. A rule is not the whole function

The domain tells you which inputs are allowed. The range is the set of outputs those inputs actually produce.

Begin with any stated domain. Then check restrictions imposed by the expression: a real square root needs a non-negative radicand, and a real logarithm needs a positive argument. Work out the range from the resulting outputs, not by copying the domain. An endpoint may be included in one set but excluded in another. When a question says “maximal real domain”, find every real input for which that expression is defined.

Watch the distinction

Do not write “all real numbers” just because a graph looks smooth. A domain restriction can come from the formula or from the question.

WORKED EXAMPLE

Find the maximal real domain and range of f(x) = √(x − 2) + 1.

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. Require x − 2 ≥ 0, so x ≥ 2. The endpoint is allowed because √0 exists.
  2. The square-root term is non-negative. Adding 1 gives f(x) ≥ 1.
  3. Domain: [2, ∞). Range: [1, ∞).

Substitute the endpoint: f(2) = 1. Large x gives arbitrarily large output.

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Now let g(x) = (x − 1)², with domain [2, 4]. Find its range.

Write your reasoning before opening the explanation.

Check your reasoning
  1. The stated domain is part of the function: do not replace it with all real numbers.
  2. For 2 ≤ x ≤ 4, x − 1 runs from 1 to 3 and its square increases from 1 to 9.
  3. Range: [1, 9].

The unrestricted parabola has minimum 0, but x = 1 is not in this domain.

2. Read transformations from the inside out

In y = a f(b(x − h)) + k, an original point (u, v) maps to (u/b + h, av + k), for non-zero b.

Use the input equation b(x − h) = u to locate the new x-coordinate. Apply the outside operations to v to get the new y-coordinate. The horizontal scale factor is 1/|b|, with a reflection in the y-axis when b is negative; the vertical scale factor is |a|, with a reflection in the x-axis when a is negative. This point mapping is safer than memorising a list of arrows, especially when several transformations are combined.

Watch the distinction

The inside factor works reciprocally. f(2x) compresses horizontally; it does not double every x-coordinate.

WORKED EXAMPLE

Describe the vertex, opening and range of y = −2(x − 3)² + 4.

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. The parent y = x² has vertex (0, 0). Translation gives the vertex (3, 4).
  2. The factor −2 reflects the graph in the x-axis and doubles vertical distances from its translated baseline. It opens downward.
  3. Domain: ℝ. Range: (−∞, 4].

At x = 3 the squared term is 0; all other squared terms make y smaller than or equal to 4.

TRY THIS NEXT

The graph of f contains (2, 5). Where does this point go on y = 3f(2(x + 1)) − 4?

Write your reasoning before opening the explanation.

Check your reasoning
  1. Solve 2(x + 1) = 2 to obtain x = 0.
  2. Transform the output: y = 3 × 5 − 4 = 11.
  3. The image of this point is (0, 11).

Use h = −1 in the general mapping. Do not read x + 1 as a shift one unit to the right.

3. Composition means the output becomes the next input

For f(g(x)), apply g first. Its output must be allowed as an input to f.

First require x to belong to the domain of g. Then require g(x) to belong to the domain of f. A simplified formula does not remove these original restrictions. The range of g need not always lie wholly inside the domain of f for a restricted composition to be described, but the relevant inputs must be selected. If a question asks whether a composite exists on the whole stated inner domain, check the full range–domain condition.

Watch the distinction

f(g(x)) and g(f(x)) generally differ. Keep the order visible before simplifying.

WORKED EXAMPLE

Let f(x) = √x on [0, ∞) and g(x) = x − 3 on ℝ. Specify f(g(x)) on its maximal allowed domain.

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. Apply g, then f: f(g(x)) = √(x − 3).
  2. Require g(x) ≥ 0, so x ≥ 3.
  3. The restricted composite has domain [3, ∞), not all of g’s original domain.

x = 2 is allowed into g but gives −1, which is not allowed into the real square-root function.

TRY THIS NEXT

For f(x) = 2x + 1 and g(x) = x² on ℝ, compare f(g(2)) and g(f(2)).

Write your reasoning before opening the explanation.

Check your reasoning
  1. g(2) = 4, so f(g(2)) = f(4) = 9.
  2. f(2) = 5, so g(f(2)) = g(5) = 25.
  3. The different results show that composition is not generally commutative.

Write the intermediate value. Squaring and doubling-plus-one cannot simply be swapped.

4. An inverse reverses a one-to-one function

An inverse function reverses inputs and outputs. The original range becomes the inverse domain.

A function must be one-to-one on its domain to have an inverse function there. On a graph, a horizontal line must not intersect it more than once. To find the inverse rule, write y = f(x), interchange x and y and solve for y, using the original restriction to choose any branch. Reflecting across y = x shows the same input–output reversal. The inverse notation f⁻¹ does not mean the reciprocal 1/f.

Watch the distinction

A square root introduces a branch choice. Do not automatically keep both signs and call the result an inverse function.

WORKED EXAMPLE

Find the inverse of f(x) = (x − 1)² with domain [1, ∞).

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. The original range is [0, ∞). On x ≥ 1, the function is one-to-one.
  2. Interchange variables: x = (y − 1)². Since original inputs satisfy y ≥ 1, choose y − 1 = √x.
  3. f⁻¹(x) = 1 + √x, with domain [0, ∞) and range [1, ∞).

f(3) = 4 and f⁻¹(4) = 3. The non-negative branch follows from the original domain.

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Why does f(x) = x² on ℝ not have an inverse function on its full range?

Write your reasoning before opening the explanation.

Check your reasoning
  1. Both 2 and −2 produce the output 4.
  2. Reversing that output would require one input, 4, to have two outputs.
  3. Restricting the original domain to [0, ∞), for example, makes an inverse function possible.

The issue is one-to-one behaviour, not whether the algebraic rule can be rearranged.

5. Exponentials and logarithms reverse one another

ln(eˣ) = x for real x, while e^(ln x) = x requires x > 0.

A logarithm asks which power gives a positive number. The argument must be positive for a real logarithm; ln(0) is not defined. Translate that argument restriction before solving equations. Exponential values are positive, even for negative exponents. Keep exact forms such as ln(5) when the question asks for an exact answer. A graph can help check a result, but a rounded calculator value is not an exact expression.

Watch the distinction

ln(a + b) is not generally ln(a) + ln(b). Logarithm laws for products do not turn sums into products.

WORKED EXAMPLE

Solve e^(2x) = 5 exactly.

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. Apply the natural logarithm to both positive sides: 2x = ln(5).
  2. Divide by 2: x = ln(5)/2.
  3. This is the exact answer; a decimal would be an approximation.

Substitution gives e^(ln 5) = 5, as required.

TRY THIS NEXT

Find the maximal real domain of h(x) = ln(3 − x), and solve h(x) = 0.

Write your reasoning before opening the explanation.

Check your reasoning
  1. Require 3 − x > 0, so the domain is (−∞, 3).
  2. ln(3 − x) = 0 means 3 − x = e⁰ = 1. Hence x = 2.
  3. x = 2 lies in the domain, so it is valid.

x = 3 is excluded: the logarithm’s argument would be zero.

6. Check the interval before finishing a circular equation

For y = a sin(bx) + k with non-zero a and b, amplitude is |a| and period is 2π/|b|.

Amplitude measures distance from the midline y = k, not the maximum y-value. Read the required interval and whether endpoints are included. A circular equation can have more than one solution in an interval, and an inverse-trig calculator output alone may not list them all. These examples use radians. The amplitude-and-period statement here is for sine; do not assign a finite amplitude to tangent.

Watch the distinction

The period changes with the inside coefficient. sin(2x) repeats after π, not after 2π.

WORKED EXAMPLE

State the amplitude, period, midline and range of y = 3 sin(2x) − 1.

Write your reasoning before opening the explanation.

Walk through the reasoning
  1. Amplitude = |3| = 3. Period = 2π/2 = π.
  2. The midline is y = −1.
  3. The range is [−1 − 3, −1 + 3] = [−4, 2].

The maximum is 2, but the amplitude is 3. These are different quantities.

TRY THIS NEXT

Solve sin x = 1/2 for 0 ≤ x ≤ 2π.

Write your reasoning before opening the explanation.

Check your reasoning
  1. The reference angle is π/6. Sine is positive in the first and second quadrants.
  2. The solutions in the interval are x = π/6 and x = 5π/6.
  3. Neither endpoint satisfies the equation.

A single inverse-sine value misses the second solution.

MAKE THE IDEA VISIBLE

Build the graph from its features

Use the controls to match three sets of graph features. Explain which parameter changed before moving on.

BUILD THE GRAPH1 / 3

Build an upward-opening parabola with vertex (2, −1) and the same vertical scale as y = x².

Change a, h and k in y = a(x − h)² + k. The graph is clipped to the displayed window; the mathematical domain is still ℝ.

-6-4-20246-6-3036912xyy = (x − 0)² + 0

Exploration only. No saved progress or XP. The target overlay is a learning aid, not an assessment shortcut.

RETRIEVE, DON’T JUST REREAD

Check the distinctions on fresh questions.

These are informal concept checks, not a predicted study score. If you need a retry, use the explanation to identify the exact mistake.

EXPLAIN IT TO YOURSELF0/6
Domain and range · CHECK 1

What is the maximal real domain of √(5 − x)?

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Keep going, with the right scope.

This guide is a starting point. Before calling this AOS revised, return to the official study design and your school’s program. Further work includes:

  • More complex combinations and parameter investigations
  • Hybrid functions and contextual modelling
  • Further power and circular-function behaviour
  • Your school’s complete course, technology requirements and assessment tasks

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