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Cambridge International AS Level · Mathematics 9709

Functions

This original practice pack develops Cambridge 9709 Pure Mathematics 1 section 1.2 skills: function language, domains and ranges, composition and its domain condition, one-one functions, inverses, inverse graphs, and standard translations, reflections and stretches. Every task is standalone and uses original values rather than copied examination material. The pack covers section 1.2 only and is not complete AS Mathematics coverage.

2026 / 2027 · Academic review not recorded · Published 2026-09-08

AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.

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17 lesson sections

Lesson contents · 17 sections

What you will learn

  • Use function, domain, range, one-one, inverse and composition terminology accurately, including mapping notation.
  • Find the range of a function in simple unrestricted and restricted-domain cases.
  • Form composite functions in the correct order and enforce the condition that the inner range lies within the outer domain.
  • Decide whether a function is one-one on its stated domain and use a suitable domain restriction when required.
  • Find inverses of simple one-one functions, state their domains and ranges, and verify them by composition.
  • Relate a one-one function and its inverse by reflection in y = x.
  • Describe and apply translations, reflections and stretches represented by y = f(x) + a, y = f(x + a), y = af(x), y = f(ax), and simple combinations.

Explanation

Functions as rules with stated inputs

A function assigns exactly one output to every input in its domain. Different inputs may share an output, but a single allowed input cannot have two outputs. In f(x) = 2x - 5, the symbol f names the function and f(4) = 3 is the image of 4. The notation f: x maps to 2x - 5 describes the same rule.

The domain is the set of allowed inputs. The range is the set of outputs actually produced from that domain. These sets are part of the function, so the same algebraic rule can define different functions when its domain changes. For example, x squared is not one-one on all real numbers but is one-one on x >= 0.

A relation fails to be a function if an allowed input has more than one image. On a graph, the vertical-line test expresses this requirement: every vertical line must meet the graph at most once. This pack uses real-valued functions unless another domain is stated.

Explanation

Domain restrictions must travel with the rule

For a formula, first identify values that make the expression meaningful. A denominator cannot be zero, and the expression inside a real square root cannot be negative. Thus 1/(x - 3) has domain x != 3, while sqrt(2x + 6) has domain x >= -3.

A question may impose a narrower domain than the formula itself requires. If f(x) = x squared for -2 <= x <= 1, those endpoints determine the attainable outputs. Never silently replace a stated domain by all real numbers, because that can change the range, whether the function is one-one, and whether an inverse exists.

Interval language should distinguish strict and inclusive boundaries. For instance, x > 2 excludes 2, while x >= 2 includes it. A reciprocal may approach zero without attaining it, so its range often has an open boundary even when its domain has a closed endpoint.

Worked example

Worked example: range of a restricted reciprocal

Let f(x) = 1/(x - 1) with domain x >= 3. Find the range. Since x - 1 >= 2, the denominator is positive and grows from 2 without bound. Therefore f(3) = 1/2 is the greatest output.

As x increases, 1/(x - 1) remains positive and decreases towards zero. It never equals zero because a reciprocal with numerator 1 cannot be zero. Hence every output y satisfies 0 < y <= 1/2.

To confirm that no values in this interval are missed, solve y = 1/(x - 1): x = 1 + 1/y. For every 0 < y <= 1/2, this gives x >= 3, so the full range is 0 < y <= 1/2.

Explanation

Finding a range systematically

A range can often be found from a familiar graph, an inequality, or by rearranging y = f(x). For a completed square, use that a real square is non-negative. For a reciprocal, consider its sign, endpoint values and unattained asymptotes. For a stated finite interval, inspect relevant endpoints and turning points.

When rearranging y = f(x), ask which y-values produce an allowed real x. This method exposes restrictions cleanly. If y = sqrt(x + 4) - 2, then y + 2 is a square root and must be non-negative, so y >= -2; conversely x = (y + 2) squared - 4 produces an allowed x for every such y.

Do not infer a range from a few sampled values. Sampling can miss an extremum or an excluded value. Instead, use the structure of the formula together with the entire stated domain, and say whether each boundary is included.

Worked example

Worked example: a quadratic range on a restricted domain

Let g(x) = (x - 1) squared + 3 for x >= 4. The unrestricted vertex is at x = 1, but that input is outside the stated domain. On x >= 4, the quantity x - 1 is positive and increasing.

The least permitted input is x = 4, giving g(4) = 3 squared + 3 = 12. As x grows, the squared term grows without bound, so there is no greatest output.

Therefore the range is g(x) >= 12. Reporting g(x) >= 3 would wrongly use the unrestricted vertex and ignore the domain. This is why domain and range should always be handled together.

Explanation

Composition means one function acts after another

The composite gf means g after f: gf(x) = g(f(x)). Work from the inside outward. In general fg and gf are different because they apply the rules in opposite orders. Substitute the complete inner expression, using brackets before simplifying.

A composite is defined only when every output under consideration from the inner function is an allowed input for the outer function. In set language, the relevant range of f must lie within the domain of g for gf to be formed on the whole stated domain of f. Otherwise the composite domain must be restricted, if the question permits it.

Formula restrictions remain active after substitution. If g(u) = sqrt(u), then gf(x) requires f(x) >= 0. If g(u) = 1/(u - 2), then gf(x) requires f(x) != 2. Finding the algebraic expression without its domain is incomplete when a restriction arises.

Worked example

Worked example: composite formula and domain

Let f(x) = 2x - 1 for real x and let g(x) = sqrt(x), whose domain is x >= 0. Then gf(x) = g(f(x)) = sqrt(2x - 1). For this square root to be real, 2x - 1 >= 0, so gf has domain x >= 1/2.

In the other order, fg(x) = f(g(x)) = 2sqrt(x) - 1. Here the inner function g already requires x >= 0, and every output of g is accepted by f, so fg has domain x >= 0.

The two composites have different formulas and domains: gf(x) = sqrt(2x - 1) for x >= 1/2, whereas fg(x) = 2sqrt(x) - 1 for x >= 0. The order cannot be inferred from ordinary multiplication.

Explanation

One-one functions and domain restriction

A function is one-one when different domain inputs always have different outputs. Equivalently, f(a) = f(b) must force a = b. Graphically, a one-one function passes the horizontal-line test: every horizontal line meets its graph at most once.

A strictly increasing or strictly decreasing function on its stated domain is one-one. A full quadratic normally is not one-one because points equally spaced from its axis have the same output. Restricting the domain to one side of the vertex removes that duplication and can make an inverse possible.

The domain restriction is not cosmetic. The restrictions x >= h and x <= h choose different branches of y = (x - h) squared + k and therefore produce different inverse formulas. Preserve the stated inequality when selecting the sign of a square root.

Worked example

Worked example: inverse of a restricted quadratic

Define h(x) = (x + 2) squared - 5 for x <= -2. This branch is strictly decreasing as x moves towards -2 and therefore is one-one. Its minimum output is -5, so its range is y >= -5.

Set y = (x + 2) squared - 5 and rearrange: (x + 2) squared = y + 5. Because the original domain gives x + 2 <= 0, choose the negative branch x + 2 = -sqrt(y + 5). Thus x = -2 - sqrt(y + 5).

Swap variable names to obtain h inverse(x) = -2 - sqrt(x + 5), with domain x >= -5 and range x <= -2. The positive square-root branch would belong to the different restriction x >= -2, so it is not valid here.

Explanation

Finding and checking an inverse

Only a one-one function has an inverse function on its stated domain. To find it algebraically, write y = f(x), rearrange to make x the subject, then interchange x and y. Record the inverse domain as the original range and the inverse range as the original domain.

A correct inverse reverses the original mapping. Consequently f inverse(f(x)) = x for every x in the domain of f, and f(f inverse(x)) = x for every x in the domain of f inverse. These identities are strong checks, but each must be used on its valid domain.

The notation f inverse does not mean 1/f(x). A reciprocal changes each output y to 1/y, while an inverse function exchanges input and output roles. For f(x) = 2x + 1, the inverse is (x - 1)/2, not 1/(2x + 1).

Worked example

Worked example: inverse of a rational function

Let f(x) = (3x - 5)/(x + 2), with x != -2. Put y = (3x - 5)/(x + 2), then multiply by x + 2: yx + 2y = 3x - 5. Collect the x-terms to get x(y - 3) = -5 - 2y.

Therefore x = (2y + 5)/(3 - y), and f inverse(x) = (2x + 5)/(3 - x). The inverse domain excludes x = 3, matching the fact that the original function can never output 3. Its range excludes -2, matching the original domain.

A composition check gives f inverse(f(x)) = x after simplifying, provided x != -2. Likewise f(f inverse(x)) = x for x != 3. The exclusions prevent division by zero and are part of the answer.

Explanation

A function and its inverse on a graph

If f maps a to b, then f inverse maps b back to a. Therefore the point (a, b) on y = f(x) becomes (b, a) on y = f inverse(x). Swapping coordinates is exactly reflection in the line y = x.

The domain and range also swap. A vertical boundary or asymptote x = c for the original corresponds to a horizontal boundary or asymptote y = c for the inverse, and vice versa. Any point on y = x stays fixed under the reflection.

A graphical inverse must still be a function, so the original graph must be one-one. If the original fails the horizontal-line test, its reflection fails the vertical-line test. Restricting the original domain to a one-one branch resolves this when appropriate.

Worked example

Worked example: reflect a cubic function to obtain its inverse

Let f(x) = (x - 1) cubed + 2. This function is strictly increasing for real x, so it is one-one. From y - 2 = (x - 1) cubed, take the real cube root to get x = 1 + cube root(y - 2).

Hence f inverse(x) = 1 + cube root(x - 2). The point (3, 10) lies on f because (3 - 1) cubed + 2 = 10, so the reflected point (10, 3) lies on the inverse.

Both graphs meet the mirror line where f(x) = x. A sketch should show y = x and place corresponding points symmetrically; it should not merely draw two unrelated curves.

Explanation

Translations: changes outside and inside f

The graph y = f(x) + a is the graph of y = f(x) translated vertically by vector (0, a). Every point (p, q) becomes (p, q + a). A positive a moves the graph up, while a negative a moves it down.

The graph y = f(x + a) is translated horizontally by vector (-a, 0). The sign appears reversed because the same original input p now occurs when x + a = p, so x = p - a. Thus f(x + 4) moves left by 4, not right.

For y = f(x + a) + b, combine the independent moves: translation by vector (-a, b). Track a known point or a vertex to verify the direction instead of relying on a memorised phrase alone.

Worked example

Worked example: translate a parabola

Start with y = f(x) = x squared, whose vertex is (0, 0). Consider y = f(x - 3) + 4 = (x - 3) squared + 4. The inside change x - 3 shifts the graph right by 3, and the outside +4 shifts it up by 4.

The combined translation vector is (3, 4), so the vertex becomes (3, 4). The point (2, 4) on y = x squared becomes (5, 8), and substitution confirms (5 - 3) squared + 4 = 8.

The shape and opening direction are unchanged because a translation does not stretch or reflect the graph. The range changes from y >= 0 to y >= 4, while the domain remains all real numbers.

Explanation

Stretches, reflections and simple combinations

For y = af(x), multiply every y-coordinate by a. If |a| > 1 this is a vertical stretch by factor |a|; if 0 < |a| < 1 it is a vertical compression. A negative a also reflects the graph in the x-axis. A point (p, q) becomes (p, aq). These stretch and reflection rules assume non-zero multipliers; if a = 0 outside the function, the graph collapses to the line y = 0.

For y = f(ax), with a != 0, divide every x-coordinate by a. The horizontal scale factor is 1/|a|. A negative a also reflects in the y-axis. A point (p, q) becomes (p/a, q). This reciprocal horizontal factor is a common source of mistakes, and coordinate division is not defined when a = 0.

For a combination, map coordinates from the original graph or solve the inner argument for the new x-coordinate. In y = A f(B(x - h)) + k, with B != 0, a base point (p, q) maps to (h + p/B, k + Aq). This point rule captures translations, reflections and stretches without ambiguity.

Worked example

Worked example: map a point through two transformations

A point (6, -4) lies on y = f(x). Find the corresponding point on y = -2f(3x). The inner argument must equal the old input: 3x = 6, so the new x-coordinate is 2.

The outside multiplier changes the old output -4 to -2(-4) = 8. Therefore the corresponding point is (2, 8). The graph has horizontal scale factor 1/3, vertical stretch factor 2, and reflection in the x-axis.

The negative sign outside affects y, not x; the 3 inside affects x, not y. Substituting the mapped coordinate confirms that -2f(3 times 2) = -2f(6) = 8.

Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.