P1 2 Functions | A Level Maths 9709

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🎯What you need to be able to do

Use the language precisely: domain, range, one-one, and the notation \( \mathrm{f}(x) \).
Find the range of a function, including where a restricted domain changes it.
Determine whether a function is one-one, and restrict a domain to make it so.
Form and evaluate composite functions, and find the domain of a composite.
Find an inverse function, state its domain and range, and sketch it as a reflection in \( y = x \).
Apply translations, reflections and stretches to a graph, and give the equation of the image.
Describe a sequence of transformations, in the correct order.

📚The mathematics

Domain, range and function notation

A function maps each input to exactly one output. The domain is the set of allowed inputs, and it is normally given to you in 9709 questions — read it, because it is doing work. The range is the set of outputs actually produced, and you have to find it.

Range questions are shape questions, so sketch first. For \( \mathrm{f}(x) = x^{2} - 4x + 7 \) on \( x \in \mathbb{R} \), completing the square gives \( (x-2)^{2} + 3 \), so the minimum is 3 and the range is \( \mathrm{f}(x) \ge 3 \). That link to completing the square is the standard route into a range at P1.

A restricted domain changes the range, and this is where care is needed. The same function on \( x \ge 4 \) has its minimum at the endpoint rather than the vertex, giving \( \mathrm{f}(x) \ge 7 \). Always check whether the vertex lies inside the given domain.

One-one functions

A function is one-one if no two different inputs give the same output — the horizontal line test. This matters because only a one-one function has an inverse.

A parabola on its full domain is not one-one, since \( \mathrm{f}(1) = \mathrm{f}(3) \) for a curve symmetric about \( x = 2 \). Restrict the domain to one side of the line of symmetry and it becomes one-one. So “state the least value of \(k\) for which \( \mathrm{f} \) is one-one on \( x \ge k \)” is asking for the \(x\)-coordinate of the vertex — a completing-the-square question wearing a disguise.

Composite functions

\( \mathrm{fg}(x) \) means \( \mathrm{f}\!\left(\mathrm{g}(x)\right) \): the function written next to the \(x\) acts first. Substitute the whole of \( \mathrm{g}(x) \) into \( \mathrm{f} \), keeping brackets.

Order matters: \( \mathrm{fg} \) and \( \mathrm{gf} \) are different functions, and questions exploit that. The composite \( \mathrm{fg} \) exists only if the range of \( \mathrm{g} \) fits inside the domain of \( \mathrm{f} \), and the domain of \( \mathrm{fg} \) is the domain of \( \mathrm{g} \) — the inner function decides what you are allowed to put in.

Inverse functions

The inverse undoes the function: \( \mathrm{f}^{-1}\mathrm{f}(x) = x \). To find it, write \( y = \mathrm{f}(x) \), rearrange to make \(x\) the subject, then swap the letters.

For a quadratic on a restricted domain, that rearrangement goes through the completed square, and the \( \pm \) that appears when you take the square root must be resolved by the domain:

\[ y = (x-2)^{2} + 3 \;\Longrightarrow\; x = 2 \pm \sqrt{y - 3} \]

If the domain was \( x \ge 2 \) take the \( + \); if \( x \le 2 \) take the \( - \). Choosing the sign is a mark, and guessing it is a coin flip.

Two facts about domain and range swap over, and they are examined directly:

domain of \( \mathrm{f}^{-1} \) = range of \( \mathrm{f} \)

range of \( \mathrm{f}^{-1} \) = domain of \( \mathrm{f} \)

Graphically, \( y = \mathrm{f}^{-1}(x) \) is \( y = \mathrm{f}(x) \) reflected in the line \( y = x \). A sketch showing both curves should show that line, dashed, and the symmetry between them — and any point where the two curves meet lies on \( y = x \), which is often the fast way to solve \( \mathrm{f}(x) = \mathrm{f}^{-1}(x) \).

\( \mathrm{f}^{-1}(x) \) is not \( \dfrac{1}{\mathrm{f}(x)} \). The superscript \( -1 \) means “inverse function”, not “reciprocal”. The same notation clash returns in trigonometry, where \( \sin^{-1}x \) means \( \arcsin x \) and not \( \dfrac{1}{\sin x} \).

Transformations of graphs

\( y = \mathrm{f}(x) + a \) — translate up \(a\)

\( y = \mathrm{f}(x + a) \) — translate left \(a\)

\( y = a\,\mathrm{f}(x) \) — stretch, factor \(a\), \(y\)-direction

\( y = \mathrm{f}(ax) \) — stretch, factor \( \tfrac{1}{a} \), \(x\)-direction

\( y = -\mathrm{f}(x) \) — reflect in the \(x\)-axis

\( y = \mathrm{f}(-x) \) — reflect in the \(y\)-axis

One rule organises all six: outside the bracket acts vertically and does what it says; inside the bracket acts horizontally and does the opposite of what it looks like. That is why \( \mathrm{f}(x+3) \) moves the graph left, and why \( \mathrm{f}(2x) \) squashes rather than stretches.

Cambridge asks for translations as a column vector, so \( \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) means 3 right and 2 down. When describing a sequence, state the transformations in order, and remember the order changes the result: a stretch then a translation is not the same as the translation then the stretch.

✏️Worked example

The function \( \mathrm{f} \) is defined by \( \mathrm{f}(x) = x^{2} - 6x + 14 \) for \( x \ge 3 \). (a) Express \( \mathrm{f}(x) \) in completed square form and state the range of \( \mathrm{f} \). (b) Explain why \( \mathrm{f} \) has an inverse, and find \( \mathrm{f}^{-1}(x) \), stating its domain. (c) The function \( \mathrm{g} \) is defined by \( \mathrm{g}(x) = 2x + 1 \) for \( x \in \mathbb{R} \). Find \( \mathrm{gf}(4) \) and solve \( \mathrm{fg}(x) = 14 \).

(a) \( x^{2} - 6x + 14 = (x-3)^{2} - 9 + 14 = (x-3)^{2} + 5 \). The domain is \( x \ge 3 \), so the bracket is never negative and takes its least value 0 at \( x = 3 \). Hence the range is \( \mathrm{f}(x) \ge 5 \).

(b) The line of symmetry is \( x = 3 \), and the domain \( x \ge 3 \) is exactly one side of it, so \( \mathrm{f} \) is one-one and therefore has an inverse. Rearranging:

\[ y = (x-3)^{2} + 5 \;\Longrightarrow\; (x-3)^{2} = y - 5 \;\Longrightarrow\; x = 3 + \sqrt{y-5} \]

The positive root is taken because the domain is \( x \ge 3 \). Swapping the letters,

\[ \mathrm{f}^{-1}(x) = 3 + \sqrt{x-5}, \qquad x \ge 5 \]

The domain of \( \mathrm{f}^{-1} \) is the range of \( \mathrm{f} \), which part (a) already gave us as \( x \ge 5 \) — no extra work needed.

(c) \( \mathrm{f}(4) = (4-3)^{2} + 5 = 6 \), so \( \mathrm{gf}(4) = \mathrm{g}(6) = 13 \).

For the equation, \( \mathrm{g} \) acts first: \( \mathrm{fg}(x) = \mathrm{f}(2x+1) = (2x + 1 - 3)^{2} + 5 = (2x-2)^{2} + 5 \). Setting this to 14:

\[ (2x-2)^{2} = 9 \;\Longrightarrow\; 2x - 2 = \pm 3 \;\Longrightarrow\; x = \tfrac{5}{2} \ \text{or}\ x = -\tfrac{1}{2} \]

But \( \mathrm{f} \) is only defined for \( x \ge 3 \), so we need \( 2x + 1 \ge 3 \), that is \( x \ge 1 \). The root \( x = -\tfrac{1}{2} \) fails this and must be rejected. The solution is \( x = \tfrac{5}{2} \).

Check it. \( \mathrm{g}\!\left(\tfrac{5}{2}\right) = 6 \), which is \( \ge 3 \) so it is a legal input to \( \mathrm{f} \), and \( \mathrm{f}(6) = 9 + 5 = 14 \). ✓ The rejected root would have required \( \mathrm{f}(0) \), and 0 is not in the domain of \( \mathrm{f} \) at all.

The restricted domain is not decoration — it is the question. It fixes the sign in part (b) and it eliminates a root in part (c). Candidates who write \( \mathrm{f}^{-1}(x) = 3 \pm \sqrt{x-5} \), or who give both roots in (c), have done all the algebra correctly and still lost the marks. Whenever a function comes with a domain, ask at every stage what that domain rules out.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. \( \mathrm{f}(x) = 2x^{2} + 8x + 3 \) for \( x \in \mathbb{R} \). Find the range, and the least value of \(k\) such that \( \mathrm{f} \) is one-one on \( x \ge k \).

Complete the square: \( 2\left(x^{2} + 4x\right) + 3 = 2\left[(x+2)^{2} - 4\right] + 3 = 2(x+2)^{2} - 5 \). The minimum is \( -5 \), so the range is \( \mathrm{f}(x) \ge -5 \). The line of symmetry is \( x = -2 \), so \( \mathrm{f} \) is one-one on any domain lying entirely to one side of it: the least such \(k\) is \( k = -2 \).

2. \( \mathrm{f}(x) = \dfrac{3}{x - 2} \) for \( x > 2 \). Find \( \mathrm{f}^{-1}(x) \) and state its domain.

Let \( y = \dfrac{3}{x-2} \). Then \( x - 2 = \dfrac{3}{y} \), so \( x = 2 + \dfrac{3}{y} \). Swapping, \( \mathrm{f}^{-1}(x) = 2 + \dfrac{3}{x} \). For the domain, find the range of \( \mathrm{f} \): as \( x \) runs over \( x > 2 \), the denominator runs over all positive values, so \( \mathrm{f}(x) > 0 \). Hence the domain of \( \mathrm{f}^{-1} \) is \( x > 0 \). Stating just \( x \ne 0 \) is not enough — the range of the original was strictly positive.

3. \( \mathrm{f}(x) = x + 3 \) and \( \mathrm{g}(x) = x^{2} \), both for \( x \in \mathbb{R} \). Find \( \mathrm{fg}(x) \) and \( \mathrm{gf}(x) \), and solve \( \mathrm{fg}(x) = \mathrm{gf}(x) \).

\( \mathrm{fg}(x) = \mathrm{f}\!\left(x^{2}\right) = x^{2} + 3 \). \( \mathrm{gf}(x) = \mathrm{g}(x+3) = (x+3)^{2} = x^{2} + 6x + 9 \). Setting them equal: \( x^{2} + 3 = x^{2} + 6x + 9 \), so \( 0 = 6x + 6 \) and \( x = -1 \). The \( x^{2} \) terms cancelling is typical — these equations are usually easier than they look.

4. The graph of \( y = x^{2} \) is translated by \( \begin{pmatrix} -2 \\ 5 \end{pmatrix} \) and then stretched by factor 3 in the \(y\)-direction. Find the equation of the resulting curve.

The translation gives \( y = (x + 2)^{2} + 5 \) — note the \( -2 \) in the vector means 2 to the left, which appears as \( +2 \) inside the bracket. The vertical stretch multiplies the whole expression by 3: \( y = 3\left[(x+2)^{2} + 5\right] = 3(x+2)^{2} + 15 \). Doing the stretch first would have given \( 3(x+2)^{2} + 5 \), which is a different curve — order matters.

5. \( \mathrm{f}(x) = (x - 1)^{2} - 4 \) for \( x \le 1 \). Find \( \mathrm{f}^{-1}(x) \) and state the range of \( \mathrm{f}^{-1} \).

\( y = (x-1)^{2} - 4 \) gives \( (x-1)^{2} = y + 4 \), so \( x = 1 \pm \sqrt{y+4} \). The domain is \( x \le 1 \), so take the negative root: \( \mathrm{f}^{-1}(x) = 1 - \sqrt{x+4} \). The range of \( \mathrm{f}^{-1} \) equals the domain of \( \mathrm{f} \), so it is \( \mathrm{f}^{-1}(x) \le 1 \). (Its domain is the range of \( \mathrm{f} \), namely \( x \ge -4 \).)

6. Describe fully the single transformation that maps \( y = \sin x \) onto \( y = \sin\!\left(\tfrac{1}{2}x\right) \).

The change is inside the function, so it acts horizontally and does the opposite of what it looks like: \( \mathrm{f}(ax) \) is a stretch of factor \( \tfrac{1}{a} \) in the \(x\)-direction, and here \( a = \tfrac{1}{2} \), giving factor 2. So it is a stretch, scale factor 2, in the \(x\)-direction. This doubles the period from \( 360^\circ \) to \( 720^\circ \), which is a useful sanity check — the graph is being pulled wider, not squashed.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Desmos — plot a function and its inverse with \( y = x \) dashed, and the symmetry becomes obvious
GeoGebra — sliders on \(a\) and \(b\) in \( a\,\mathrm{f}(x+b) \) to feel the four transformations
Khan Academy — composite and inverse functions