9709 Pure 1 · 1.2 Functions · Session 3 of 5

Running a Function Backwards

A function takes an input and gives an output. Its inverse takes that output and hands back the input you started with. Some functions can be run backwards and some can't.

Do Now: finish Tuesday's exam question5 min
What an inverse does, and when one exists10 min
Finding \(f^{-1}(x)\), two ways, on two real exam questions14 min
The square-root sign trap5 min
The graph: the mirror line \(y = x\), and the Mirror Lab8 min
Practice, then the exit ticket15 min

Do Now · 5 min · Specimen Paper 1, Q2 · the last step

Finish Tuesday's question. Tap to check.

\(f : x \mapsto 3x + 2\) and \(g : x \mapsto 4x - 12\), both for \(x \in \mathbb{R}\). Solve the equation \(f^{-1}(x) = gf(x)\). [4]

On Tuesday we found both pieces: \(f^{-1}(x) = \tfrac{x - 2}{3}\) and \(gf(x) = 12x - 4\).

Set them equal and solve for \(x\).
write the equation, this line earns the method mark:  \(\dfrac{x - 2}{3} = 12x - 4\)
\(\times\, 3\) to clear the fraction:  \(x - 2 = 36x - 12\)
collect the \(x\) terms on the right:  \(-2 + 12 = 36x - x\)  →  \(10 = 35x\)
\(x = \dfrac{10}{35} = \dfrac{2}{7}\)

Mark scheme: B1 for \(f^{-1}(x)\), B1 for \(gf(x)\), M1 for setting them equal, A1 for \(\tfrac{2}{7}\). Today is about where that \(f^{-1}\) came from, and when it exists at all.

Direct instruction · 1 of 9

An inverse undoes the function.

\(f(x) = 2x + 1\)

Forwards: \(f\)

\(4\) → double it → \(8\) → add 1 → \(9\)
\(f(4) = 9\)

Backwards: \(f^{-1}\)

\(9\) → subtract 1 → \(8\) → halve it → \(4\)
\(f^{-1}(9) = 4\)

The inverse does the opposite operations in the opposite order. As a formula, \(f^{-1}(x) = \tfrac{x - 1}{2}\).

The \(-1\) is not a power. \(f^{-1}(9)\) means "run \(f\) backwards on 9", which is 4. It does not mean \(\tfrac{1}{f(9)}\), which would be \(\tfrac{1}{19}\). The notation looks like \(x^{-1} = \tfrac{1}{x}\), and that look-alike trips people up.

Direct instruction · 2 of 9

Only a one-one function can be run backwards.

A function is one-one when every output comes from exactly one input. It is many-one when two or more inputs share an output.

One-one · \(f(x) = 2x + 1\)

Ask "which input gave 9?" and there is one answer: 4. Any horizontal line crosses the graph once.

Many-one · \(f(x) = x^2\) for \(x \in \mathbb{R}\)

\(f(3) = 9\) and \(f(-3) = 9\). Ask "which input gave 9?" and there are two answers. A function can only give one output, so there is no inverse.

Two tests, two different questions. The vertical line test asks "is this a function at all?" Both graphs pass it: each \(x\) gives one \(y\). The horizontal line test asks "is it one-one, so does it have an inverse?" The line passes it. \(x^2\) on all of \(\mathbb{R}\) fails it, so \(x^2\) is a function but it is not one-one.

Direct instruction · 3 of 9

Cut the domain down until it passes the test.

\(x^2\) fails on all of \(\mathbb{R}\). Keep only one side of the vertex and every output comes from one input again.

\(f(x) = x^2\) for \(x \ge 0\)

Right half only. 9 now comes from 3 alone. The inverse hands back the original inputs, and those were all \(\ge 0\), so it must give the positive root.

\(f^{-1}(x) = \sqrt{x}\)
\(f(x) = x^2\) for \(x \le 0\)

Left half only. 9 now comes from \(-3\) alone. The original inputs were all \(\le 0\), so the inverse must hand back a negative number: the negative root. \(f^{-1}(9) = -3\).

\(f^{-1}(x) = -\sqrt{x}\)

This is why Cambridge functions come with "for \(x > \dots\)" attached. The restriction is what makes an inverse possible, and it decides which answer is right. You'll see that on the sign-trap slide.

Worked example · real past paper · 9709/12 June 2024, Q4(b)

\(g(x) = \dfrac{1}{x^2 + 2}\) for \(x \in \mathbb{R}\). State the range of \(g\) and explain whether \(g^{-1}\) exists. [2]

1 · the smallest \(x^2\) can be is 0, at \(x = 0\):  so the smallest bottom is \(x^2 + 2 = 2\)
2 · smallest bottom gives the biggest fraction:  \(g(0) = \dfrac{1}{2}\), the top of the range, and it is reached, so \(\le\)
3 · as \(x\) grows, the bottom grows without limit:  \(g(10) = \dfrac{1}{102}\), \(g(100) = \dfrac{1}{10002}\), getting closer to 0
4 · 1 over a positive number is never 0 or negative:  so 0 is never reached, and that end is \(>\)
range: \(0 < g(x) \le \tfrac12\)
now look for two inputs with the same output:  \(g(1) = \dfrac{1}{3}\) and \(g(-1) = \dfrac{1}{3}\)
\(g^{-1}\) does not exist, because \(g\) is not one-one.

Examiners saw many answers give only one side of the range, or mix up \(\ge\) and \(>\). Both ends count. For the reason, "not one-one", "many-one" and "fails the horizontal line test" all score.

Direct instruction · 4 of 9 · finding \(f^{-1}(x)\)

Two routes to \(f^{-1}(x)\). Same answer, both full marks.

\(f(x) = 5x - 3\)  write \(y = f(x)\) first, whichever route you take

Route 1 · swap first, then solve

\(y = 5x - 3\)
swap \(x\) and \(y\):  \(x = 5y - 3\)
\(+\,3\):  \(x + 3 = 5y\)
\(\div\, 5\):  \(y = \dfrac{x + 3}{5}\)

This is the way we did it on Tuesday. The swap happens first, so the last line is already in \(x\).

Route 2 · solve first, then swap

\(y = 5x - 3\)
\(+\,3\):  \(y + 3 = 5x\)
\(\div\, 5\):  \(x = \dfrac{y + 3}{5}\)
swap the letters:  \(y = \dfrac{x + 3}{5}\)

Make \(x\) the subject, then rename the letters. The algebra is identical. Only the moment of the swap moved.

\(f^{-1}(x) = \dfrac{x + 3}{5}\)

Pick either and stick with it. The 2023 mark scheme lists both. The one thing both need: the final line is written as \(f^{-1}(x) = \dots\) in terms of \(x\). Also swapped: domain of \(f^{-1}\) = range of \(f\), and range of \(f^{-1}\) = domain of \(f\).

Worked example · real past paper · 9709/11 June 2023, Q8(b)

\(f(x) = 1 + \dfrac{6}{x - 3}\) for \(x > 3\). Find the domain of \(f^{-1}\). [1]

Tuesday's function. The paper gave \(f(x) = 1 + \tfrac{2a}{x - a}\), and we found \(a = 3\).

the domain of \(f^{-1}\) is the range of \(f\), so find the range, starting from the domain:
1 · start from the domain:  \(x > 3\)
2 · subtract 3 from both sides:  \(x - 3 > 0\), so the bottom is always positive
3 · 6 over a positive number is positive:  \(\dfrac{6}{x - 3} > 0\)
4 · add 1 to both sides:  \(1 + \dfrac{6}{x - 3} > 1\)
range of \(f\):  \(f(x) > 1\)
domain of \(f^{-1}\):  \(x > 1\)

On the real paper, examiners saw \(x > 3\) (the domain of \(f\)), \(x \ne 1\), and range and domain mixed up. The mark scheme gives zero for \(f^{-1}(x) > 1\). A domain is a set of inputs, so write it in \(x\).

Worked example · real past paper · 9709/11 June 2023, Q8(c)

Same \(f\). Find an expression for \(f^{-1}(x)\). [3]

\(f(x) = 1 + \dfrac{6}{x - 3}\) for \(x > 3\)
write \(y = f(x)\), then swap \(x\) and \(y\):  \(x = 1 + \dfrac{6}{y - 3}\)
\(-\,1\), then flip both sides over and \(\times\, 6\):  \(x - 1 = \dfrac{6}{y - 3}\)  →  \(y - 3 = \dfrac{6}{x - 1}\)
\(+\,3\):  \(y = 3 + \dfrac{6}{x - 1}\)
\(f^{-1}(x) = 3 + \dfrac{6}{x - 1}\)  or  \(\dfrac{3x + 3}{x - 1}\)
check it: on Tuesday \(f(5) = 4\), so \(f^{-1}\) should send 4 back to 5:  \(3 + \dfrac{6}{4 - 1} = 3 + 2 = 5\) ✓

Either form earns all 3 marks. A last line still in \(y\) loses the final mark.

Direct instruction · 5 of 9 · the syllabus's own example

A square root gives \(\pm\). The domain picks one.

\(h(x) = (2x + 3)^2 - 4\) for \(x < -\dfrac{3}{2}\)

Swap and solve

\(y = (2x + 3)^2 - 4\)
swap:  \(x = (2y + 3)^2 - 4\)
\(+\,4\):  \((2y + 3)^2 = x + 4\)
square root:  \(2y + 3 = \pm\sqrt{x + 4}\)

Two possible signs. Only one of them is right.

Let the domain decide

The outputs of \(h^{-1}\) are the inputs of \(h\), so \(y < -\tfrac{3}{2}\).

\(y < -\dfrac{3}{2}\)  →  \(2y < -3\)  →  \(2y + 3 < 0\)

\(2y + 3\) is negative, so take the negative root.

\(2y + 3 = -\sqrt{x + 4}\)  →  \(2y = -3 - \sqrt{x + 4}\)
\(h^{-1}(x) = \dfrac{-3 - \sqrt{x + 4}}{2}\)  for \(x > -4\)
where \(x > -4\) comes from: domain of \(h^{-1}\) = range of \(h\).  \(x < -\tfrac{3}{2}\) → \(2x + 3 < 0\) → \((2x + 3)^2 > 0\) → \((2x + 3)^2 - 4 > -4\), so \(h(x) > -4\)

Check: \(h^{-1}(0) = \tfrac{-3 - 2}{2} = -\tfrac{5}{2}\), and \(h\big(-\tfrac{5}{2}\big) = (-2)^2 - 4 = 0\) ✓

Direct instruction · 6 of 9 · the graph

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

Finding the inverse swapped \(x\) and \(y\). So every point \((a, b)\) on \(y = f(x)\) becomes the point \((b, a)\) on \(y = f^{-1}(x)\), and swapping the coordinates is exactly what a reflection in \(y = x\) does.

\(f(4) = 9\), so \((4, 9)\) is on \(f\)  →  \((9, 4)\) is on \(f^{-1}\)

What a Cambridge sketch needs

1The line \(y = x\), drawn and labeled. The syllabus says sketches should include the mirror line \(y = x\).

2Both curves, mirror images across that line. End points and asymptotes swap too: an asymptote \(x = 3\) on \(f\) becomes \(y = 3\) on \(f^{-1}\).

3\(f^{-1}\) sitting where its domain says. Its \(x\)-values are the range of \(f\).

June 2024 examiner report: "The mirror line of \(y = x\) was often omitted." Draw the line first, then reflect.

Direct instruction · 7 of 9 · the Mirror Lab

Drag the point. Watch its reflection.

Try the last one, \(x^2\) on all of \(\mathbb{R}\). Its reflection fails: one input would need two outputs. That is the many-one problem from earlier, seen on the graph.

Direct instruction · 8 of 9 · check your answer

How to check your \(f^{-1}\) is right.

Why: \(f^{-1}\) undoes \(f\), so doing one then the other must give back exactly what you started with. If it doesn't, your inverse has a mistake. Cambridge also asks you to show this directly.

\(f(x) = 2x + 1\)  ·  \(f^{-1}(x) = \dfrac{x - 1}{2}\)
\(f^{-1}f(x)\), \(f\) first:
\(f^{-1}(2x + 1) = \dfrac{(2x + 1) - 1}{2} = \dfrac{2x}{2}\)
\(= x\)
\(ff^{-1}(x)\), \(f^{-1}\) first:
\(f\Big(\dfrac{x - 1}{2}\Big) = 2 \cdot \dfrac{x - 1}{2} + 1 = x - 1 + 1\)
\(= x\)

The quick version for exams: pick an \(x\) in the domain, work out \(f\) of it, then put that answer into your \(f^{-1}\). If you don't get your \(x\) back, the inverse is wrong. It takes thirty seconds and it catches sign errors.

Practice · together · 7 min

Three different shapes. Tap to check.

1.  Is \(p(x) = x^2 - 4x\) for \(x \in \mathbb{R}\) one-one? Justify.
\(p(0) = 0\) and \(p(4) = 16 - 16 = 0\): two inputs, one output
No. \(p\) is many-one, so \(p^{-1}\) does not exist.
2.  \(q(x) = 5 - 2x\) for \(x \in \mathbb{R}\). Find \(q^{-1}(x)\).
swap:  \(x = 5 - 2y\)  →  \(2y = 5 - x\)
\(q^{-1}(x) = \dfrac{5 - x}{2}\)
check:  \(q(1) = 3\), and \(q^{-1}(3) = \dfrac{5 - 3}{2} = 1\) ✓
3.  \(r(x) = \dfrac{4}{x + 1}\) for \(x > 0\). Find the domain of \(r^{-1}\).
\(x > 0\)  →  \(x + 1 > 1\)  →  \(0 < \dfrac{4}{x + 1} < 4\)
range of \(r\):  \(0 < r(x) < 4\)
domain of \(r^{-1}\): \(0 < x < 4\)

Practice · on your own · 8 min

Your turn. Tap to check when you're done.

4.  Same \(r\) as Q3. Find \(r^{-1}(x)\).
swap:  \(x = \dfrac{4}{y + 1}\)  →  \(y + 1 = \dfrac{4}{x}\)
\(r^{-1}(x) = \dfrac{4}{x} - 1\)
check:  \(r(1) = 2\), and \(r^{-1}(2) = 2 - 1 = 1\) ✓
5.  \(s(x) = (x - 3)^2 + 2\) for \(x \ge 3\). Find \(s^{-1}(x)\) and its domain.
swap:  \(x = (y - 3)^2 + 2\)  →  \(y - 3 = \pm\sqrt{x - 2}\)
\(y \ge 3\), so \(y - 3 \ge 0\): take the \(+\) root
\(s^{-1}(x) = 3 + \sqrt{x - 2}\) for \(x \ge 2\)
6.  The point \((2, 7)\) is on \(y = t(x)\), and \(t\) is one-one. Name a point that must be on \(y = t^{-1}(x)\).
\((7, 2)\): swap the coordinates, which reflects \((2, 7)\) in \(y = x\).

Q3 and Q4 share a function on purpose: the domain you found in Q3 is the domain of the answer to Q4.

Direct instruction · 9 of 9 · recap

Three things to carry out of here.

Only one-one functions have inverses. Two inputs with the same output means no inverse, and a restriction like "for \(x \ge 3\)" is often what makes a function one-one.

Swap, solve, then check the sign. After a square root, the domain of the original function tells you which sign to keep. Domain of \(f^{-1}\) = range of \(f\).

Sketch with the mirror line. \(f^{-1}\) is \(f\) reflected in \(y = x\). Draw and label \(y = x\) every time.

Exit ticket · 5 min

Schoology, Paper 1 → 1.2 Functions.

Four questions. One of them asks for the full method, so show how you chose the sign.

Press Submit before you close the tab. An attempt left open scores nothing.

Homework · and where to practice more

Before next class.

Next session: transformations. Moving, stretching and reflecting whole graphs.