9709 Pure 1 · 1.2 Functions · Session 2 of 5

Functions That Feed Each Other

One function's output becomes another function's input. Get the order wrong and you get a different answer, every time.

Do Now: same two functions, two different orders5 min
What a composite function is, and the two ways to write one10 min
Turning a composite into a formula6 min
When a composite can't be formed, and the slider that shows it9 min
Two real exam questions12 min
Practice, then the exit ticket13 min

Do Now · 5 min

Same two functions, two different orders.

\(f(x) = 2x - 3\)  ·  \(g(x) = 1 - 2x\)  ·  both \(x \in \mathbb{R}\)

Start from \(x = 2\) each time. Tap to check.

A.  Work out \(f\) first, then feed that result into \(g\).
\(f(2) = 2(2) - 3 = 1\)  →  \(g(1) = 1 - 2(1) = -1\)
B.  Work out \(g\) first, then feed that result into \(f\).
\(g(2) = 1 - 2(2) = -3\)  →  \(f(-3) = 2(-3) - 3 = -9\)

Two different answers from the same starting number. The order you apply the functions in changes the result.

Direct instruction · 1 of 5

Feed one function's output into another.

A composite function takes the output of one function and uses it as the input of another. Nothing new is being calculated. You're chaining two rules you already know.

Think of an assembly line

Station 1 takes a part and does something to it. Station 2 takes whatever comes off station 1 and does something else to it. Swap the order of the stations and a different product comes off the line, even though both stations do exactly the same job either way.

\(f\) then \(g\) gave \(-1\). \(g\) then \(f\) gave \(-9\). Same functions, same start, different order.

Direct instruction · 2 of 5 · notation

Two ways to write it. You'll see both.

Cambridge exam notation

\(gf(x)\)

No circle, the letters just sit next to each other. \(f\) is written right next to \(x\), so \(f\) is applied first. Then \(g\) is applied to that result.

\(gf(x) = g\big(f(x)\big)\)

Khan Academy notation

\((f \circ g)(x)\)

Read "f composed with g". \(g\) is written right next to \(x\), inside the brackets, so \(g\) is applied first. Then \(f\) is applied to that result.

\((f \circ g)(x) = f\big(g(x)\big)\)

The rule that works for both: find whichever function is written closest to \(x\) and do that one first, then apply the other to the result. Don't just read left to right. In both notations, the function you meet first on the page is applied second.

On a Cambridge paper, always answer in \(gf(x)\) form. That's what the mark scheme expects.

Direct instruction · 3 of 5 · worked example

Turn it into a formula that works for any \(x\).

\(f(x) = 2x - 3\)  ·  \(g(x) = 1 - 2x\)
\(gf(x)\): apply \(f\) first, but keep it as \(x\), no number yet:  \(gf(x) = g\big(f(x)\big) = g(2x - 3) = 1 - 2(2x - 3)\)
\(gf(x) = 1 - 4x + 6 = 7 - 4x\)
check against Row A of the Do Now, \(x = 2\):  \(7 - 4(2) = 7 - 8 = -1\) ✓
\(fg(x)\): apply \(g\) first:  \(fg(x) = f\big(g(x)\big) = f(1 - 2x) = 2(1 - 2x) - 3 = 2 - 4x - 3\)
\(fg(x) = -4x - 1\)
check against Row B, \(x = 2\):  \(-8 - 1 = -9\) ✓

Direct instruction · 4 of 5 · the condition

\(gf\) can't always be formed.

\(gf\) can only be formed when the whole range of \(f\) fits inside the domain of \(g\).

\(g\) has to be able to accept every single output \(f\) could ever produce. If \(f\) can output a number \(g\) isn't defined for, the chain breaks at that point.

See it break, with real numbers

\(f(x) = x^2\) for \(x \ge 0\)  ·  \(g(x) = \dfrac{1}{x - 5}\) for \(x > 5\)
try \(x = 1\):  \(f(1) = 1^2 = 1\).  Now \(g(1)\): is \(1 > 5\)? No. \(g(1)\) doesn't exist.

One input sent \(f\)'s output somewhere \(g\) can't follow. That's enough on its own: \(gf\) cannot be formed for this \(f\) and \(g\). The next slide lets you drag \(f\)'s domain and watch the verdict flip.

This exact condition shows up on real Cambridge papers with different letters. On the June 2025 paper most candidates missed it. The ones who got credit named exactly which range wasn't inside which domain, the same move you just made.

Direct instruction · 4 of 5 · see it flip

Drag the boundary. Watch the verdict change.

\(f(x) = x^2\) for \(x \ge\) 3.0  ·  \(g(x) = \dfrac{1}{x - 5}\) for \(x > 5\)

Drag \(k\) below about 2.24 (\(\sqrt{5}\)) and the range of \(f\) dips to include numbers \(\le 5\), which \(g\) can't accept, and the verdict flips to red. Nothing about \(g\) changed. Only \(f\)'s domain did, which changed \(f\)'s range. That's the whole condition in one slider.

Worked example · real past paper · Specimen Paper 1, Q2

\(f : x \mapsto 3x + 2\), \(g : x \mapsto 4x - 12\), both for \(x \in \mathbb{R}\). Find \(gf(x)\).

The full question asks you to solve \(f^{-1}(x) = gf(x)\). Today: find \(gf(x)\), the piece you already have the tools for.

apply \(f\) first, because \(f\) is closest to \(x\) in \(gf(x)\):  \(gf(x) = g\big(f(x)\big) = g(3x + 2)\)
put \((3x + 2)\) into \(g\)'s rule and multiply out:  \(4(3x + 2) - 12 = 12x + 8 - 12\)
\(gf(x) = 12x - 4\)

That's \(gf(x)\) done. Next, a quick preview of how \(f^{-1}\) itself is found, so that equation isn't a black box.

Preview · Session 3

How \(f^{-1}\) is found, quickly.

\(f(x) = 3x + 2\)

Full inverse-function theory, including when an inverse exists at all, is Session 3. Here's just the mechanic, so today's equation isn't a mystery.

write \(y = f(x)\), then swap \(x\) and \(y\):  \(y = 3x + 2\)  →  \(x = 3y + 2\)
solve that new equation for \(y\):  \(x - 2 = 3y\)  →  \(y = \dfrac{x - 2}{3}\)
\(f^{-1}(x) = \dfrac{x - 2}{3}\)

Session 3 covers why swapping \(x\) and \(y\) works, when an inverse exists at all, and finishes solving \(f^{-1}(x) = gf(x)\), that is \(\tfrac{x - 2}{3} = 12x - 4\).

Worked example · real past paper · 9709/11 June 2023, Q8(a) · 1 of 2

Use a composite to find missing constants.

\(f(x) = 1 + \dfrac{2a}{x - a}\) for \(x > a\), and \(g(x) = bx - 2\). Given that \(f(7) = \tfrac{5}{2}\) and \(gf(5) = 4\), find \(a\) and \(b\). [4]

use \(f(7) = \tfrac{5}{2}\) first, because it only involves \(a\):  \(1 + \dfrac{2a}{7 - a} = \dfrac{5}{2}\)  →  \(\dfrac{2a}{7 - a} = \dfrac{3}{2}\)
cross-multiply:  \(4a = 3(7 - a) = 21 - 3a\)
\(7a = 21\)  →  \(a = 3\)

With \(a\) known, the second fact, \(gf(5) = 4\), is what finds \(b\).

Worked example · real past paper · 9709/11 June 2023, Q8(a) · 2 of 2

Now use the composite to find \(b\).

\(f(x) = 1 + \dfrac{6}{x - 3}\)  ·  \(g(x) = bx - 2\)  ·  \(gf(5) = 4\)
find \(f(5)\) first, using \(a = 3\):  \(f(5) = 1 + \dfrac{6}{5 - 3} = 1 + 3 = 4\)
\(gf(5) = g\big(f(5)\big) = g(4)\):  \(g(4) = 4b - 2\)
set it equal to the given value:  \(4b - 2 = 4\)  →  \(4b = 6\)
\(b = \dfrac{3}{2}\)

Examiners noted that students who found \(f(5)\) first did best. A common slip was dropping the "1 +" from \(f\).

Practice · together · 6 min

Different shapes, on purpose. Tap to check.

1.  \(f(x) = x + 4\), \(g(x) = x^2\). Find \(gf(3)\).
\(f(3) = 7\)  →  \(g(7) = 49\)
\(gf(3) = 49\)
2.  \(f(x) = 5 - x\), \(g(x) = 2x + 1\). Find \(gf(x)\).
\(gf(x) = 2(5 - x) + 1 = 10 - 2x + 1\)
\(gf(x) = 11 - 2x\)
3.  Same \(f\) and \(g\) as Q2. Find \(fg(x)\) and compare.
\(fg(x) = 5 - (2x + 1)\)
\(fg(x) = 4 - 2x\)
different from \(gf(x) = 11 - 2x\): order matters

Practice · on your own · 7 min

Can \(gf\) be formed? Tap to check.

4.  \(f(x) = \dfrac{1}{x}\) for \(x > 0\), so \(f(x) > 0\). \(g(x) = x - 3\) for \(x \in \mathbb{R}\).
\(g\)'s domain is all of \(\mathbb{R}\), so it accepts every possible output of \(f\)
Yes, \(gf\) can be formed.
5.  \(f(x) = x^2\) for \(x \in \mathbb{R}\), so \(f(x) \ge 0\). \(g(x) = \sqrt{x - 10}\) for \(x \ge 10\).
range of \(f\) starts at 0, domain of \(g\) starts at 10. \(f(2) = 4\), and \(g\) isn't defined at 4
No, \(gf\) cannot be formed.

Direct instruction · 5 of 5 · recap

Two things to leave with.

Notation. Whichever symbol you see, \(gf(x)\) or \((f \circ g)(x)\), find the function touching \(x\) and apply that one first. On a Cambridge paper, answer in \(gf(x)\) form.

The existence condition. \(gf\) only exists when the whole range of \(f\) fits inside the domain of \(g\). Check the two sets, and name them in your answer. That's what the mark scheme rewards.

Next session: one-one functions and inverses in full, including finishing today's Specimen Paper question.

Exit ticket · 5 min

Schoology, Paper 1 → 1.2 Functions.

Four questions, each with a fresh pair of functions.

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