Functions

• A function is a rule with a domain. Ask “on which domain?” before every range, every composite and every inverse — and track one point through every transformation.
• gf means f first, and exists only if the range of f fits the domain of g. The domain of the inverse is the range of the function, and the domain picks the square root. Outside the bracket acts on y as it reads; inside acts on x the other way.

• 1.2.1 I can understand the terms function, domain, range, one-one function, inverse function and composition of functions — including that \(gf\) can only be formed when the range of \(f\) is within the domain of \(g\)
• 1.2.2 I can identify the range of a given function in simple cases, and find the composition of two given functions
• 1.2.3 I can determine whether or not a given function is one-one, and find the inverse of a one-one function in simple cases
• 1.2.4 I can illustrate in graphical terms the relation between a one-one function and its inverse, with the mirror line \(y = x\) indicated
• 1.2.5 I can understand and use the transformations of the graph of \(y = f(x)\) given by \(y = f(x) + a\), \(y = f(x + a)\), \(y = af(x)\), \(y = f(ax)\) and simple combinations of these, naming each as a translation, reflection or stretch

Why the domain now matters. The rule \(x \mapsto x^2\) has range \(y \ge 0\) on all real numbers and range \(y \ge 9\) on \(x \ge 3\); it is not one-one on the first domain and is one-one on the second. Same rule, different function. Every question in this chapter starts by asking which domain you are on.

• “The range of \(g(x) = x^2 + 1\) is \(x \ge 1\).” Correct A range is a set of output values, so it is written in \(g(x)\) or \(y\): \(g(x) \ge 1\). An inequality in \(x\) describes a domain.
• “\(y = f(x + 3)\) is the graph moved 3 units to the right.” Correct It moves 3 units to the left: the value \(f(0)\) now happens at \(x = -3\). It is a translation by the vector \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\).
• “\(gf(x)\) means do \(g\) first.” Correct The function written next to \(x\) acts first: \(gf(x) = g(f(x))\). And \(gf\) can only be formed when the range of \(f\) is within the domain of \(g\).
• “\(f(x) = (x - 3)^2 - 7\) for \(x \ge 3\) has inverse \(3 \pm \sqrt{x + 7}\).” Correct An inverse is a function, so it gives one value. The domain \(x \ge 3\) means \(x - 3 \ge 0\), which selects the positive root: \(f^{-1}(x) = 3 + \sqrt{x + 7}\).
• “\(f^{-1}(x) = \dfrac{1}{f(x)}\).” Correct The index \(-1\) on a function name means the inverse function, not the reciprocal. For \(f(x) = 2x + 1\), \(f^{-1}(x) = \dfrac{x - 1}{2}\), while \(\dfrac{1}{f(x)} = \dfrac{1}{2x + 1}\).
• “\(y = f(2x)\) is a stretch with scale factor 2.” Correct A stretch parallel to the \(x\)-axis with scale factor \(\tfrac{1}{2}\): every \(x\)-coordinate is halved.
• “The range of \((x - 2)^2 + 1\) for \(x \ge 4\) is \(f(x) \ge 1\), from the vertex.” Correct The vertex \(x = 2\) is not in the domain, so it gives no value. The least value is at the endpoint: \(f(4) = 5\), so \(f(x) \ge 5\).
• “The range of \(g(x) = x^2 + 1\) is \(x \ge 1\).” Repair A range is a set of output values: \(g(x) \ge 1\), or \(y \ge 1\).
• “\(f(x) = \tfrac{1}{x}\) for \(x \ge 1\) has range \(f(x) > 0\).” Repair The endpoint \(x = 1\) is included and gives \(f(1) = 1\), the greatest value, so the range is \(0 < f(x) \le 1\).
• “\((x - 2)^2 + 1\) for \(x \ge 4\) has range \(f(x) \ge 1\).” Repair The vertex \(x = 2\) is not in the domain; the least value is at the endpoint, \(f(4) = 5\), so \(f(x) \ge 5\).
• “\(gf(x)\) means do \(g\) first”, or “\(gf(x) = g(x) \times f(x)\)”. Repair The function nearest \(x\) acts first: \(gf(x) = g(f(x))\), \(f\) first. It is a composite, not a product.
• “\(gf\) exists for any \(f\) and \(g\).” Repair \(gf\) can only be formed when the range of \(f\) is within the domain of \(g\). With \(g(x) = \sqrt{x}\), \(x \ge 0\), and \(f(x) = x - 4\) on \(\mathbb{R}\), it cannot: \(f(0) = -4\).
• “\(f(x) = x^2\) on \(\mathbb{R}\) is one-one because each \(x\) gives one value.” Repair That is the definition of a function. One-one needs each output to come from one input, and \(f(-2) = f(2) = 4\) breaks it.
• “For \(f(x) = 2x + 1\), \(f^{-1}(x) = \dfrac{1}{2x + 1}\).” Repair \(f^{-1}\) is the inverse function, \(\dfrac{x - 1}{2}\); the reciprocal \(\dfrac{1}{f(x)}\) is a different thing.
• “\(f(x) = (x - 3)^2 - 7\), \(x \ge 3\), has inverse \(3 \pm \sqrt{x + 7}\).” Repair An inverse is a function and gives one value; \(x \ge 3\) selects the positive root, \(3 + \sqrt{x + 7}\). For the syllabus’s \(h\) on \(x < -\tfrac{3}{2}\), the same reasoning selects the negative root.
• “The domain of \(f^{-1}\) is the domain of \(f\).” Repair The domain of \(f^{-1}\) is the range of \(f\), and its range is the domain of \(f\).
• “The graphs of \(f\) and \(f^{-1}\) always meet on \(y = x\), so solve \(f(x) = x\).” Repair True for an increasing function, such as the restricted quadratics in figures 3, 4 and 7. Not every restricted quadratic is increasing: the syllabus’s \(h(x) = (2x + 3)^2 - 4\) for \(x < -\tfrac{3}{2}\) lies to the left of its vertex and is decreasing. A decreasing function can meet its inverse off the line (\(\tfrac{1}{x}\), \(x > 0\), is its own inverse), so say why you are using \(f(x) = x\).
• “\(y = f(x + 3)\) is the graph moved 3 to the right.” Repair Left: \(f(x + 3)\) takes the value \(f(0)\) at \(x = -3\). Translation by \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\).
• “\(y = f(2x)\) is a stretch with scale factor 2.” Repair Scale factor \(\tfrac{1}{2}\), parallel to the \(x\)-axis: the point \((4, q)\) goes to \((2, q)\).
• “\(y = 2f(x) + 3\): translate by \(\begin{pmatrix} 0 \\ 3 \end{pmatrix}\), then stretch by factor 2.” Repair That produces \(2(f(x) + 3) = 2f(x) + 6\). Stretch first, then translate by \(\begin{pmatrix} 0 \\ 3 \end{pmatrix}\).
• “\(y = -f(x)\) is a reflection in the \(y\)-axis.” Repair \(-f(x)\) changes the \(y\)-coordinates: reflection in the \(x\)-axis. \(f(-x)\) is the reflection in the \(y\)-axis.
• “A stretch parallel to the \(y\)-axis” — with no scale factor, or “a shift of 3” with no direction. Repair A full description is the word and its data: the vector of a translation; the direction and the scale factor of a stretch; the line of a reflection.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• The composite function \(gf\) can only be formed when the range of \(f\) is within the domain of \(g\). This is the syllabus’s own condition. With \(f : x \mapsto 2x - 5\) for \(x \in \mathbb{R}\) and \(g : x \mapsto \sqrt{x}\) for \(x \ge 0\): the range of \(f\) is all of \(\mathbb{R}\), which is not within \(x \ge 0\) (for example \(f(0) = -5\), and \(\sqrt{-5}\) is not defined), so \(gf\) cannot be formed. It can be formed once the domain of \(f\) is restricted to \(x \ge \tfrac{5}{2}\), when the range of \(f\) becomes \(f(x) \ge 0\). Worked example 2 does this in full.
• The graph of \(y = f^{-1}(x)\) is the reflection of the graph of \(y = f(x)\) in the line \(y = x\). The syllabus says a sketch should include an indication of the mirror line \(y = x\).
• Interleave with the chapters that use this one. When you reach chapter 5 (Trigonometry), re-answer worked example 7 with a trigonometric graph in place of f, and say why the inverse sine needs a restricted domain. At chapter 10 (Paper 3), re-draw figure 4 with the exponential function and its inverse. At chapter 14 (numerical solution of equations), rearrange an equation into the form x = F(x) and state the domain on which F is defined. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• The syllabus asks you to identify ranges, find composites and inverses, determine whether a function is one-one, illustrate a function with its inverse, and use transformations. A structured question can run these in sequence on one function — complete the square, state the range, choose a domain that makes it one-one, find the inverse, sketch both — so each later part leans on an earlier answer. A transformation part may give only a graph’s features (a maximum point, an intercept) and ask where they go.
• MF19 prints no result from section 1.2. The definitions, the rule that the domain of \(f^{-1}\) is the range of \(f\), the condition for \(gf\) to exist and the four transformation rules must all be known. See the MF19 card.
• Answers here are exact: surds and fractions such as \(\tfrac{-3 - \sqrt{x + 4}}{2}\), \(\tfrac{5}{3}\) and \(\tfrac{7 + \sqrt{41}}{2}\) stay as they are, and a decimal (to 3 significant figures) is used only to place a point on a sketch. A range is written in \(f(x)\) or \(y\), with \(\ge\) or \(>\) chosen by whether the endpoint is attained. A show-that on \(ff(x) = x\) needs every algebraic step visible.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.