IB Maths AI SL Topic 2 — Further Functions & Graphs Paper 1 & 2 Reflection in y = x ~7 min read

Inverse Functions

An inverse function f⁻¹(x) undoes what f(x) does — so if f(2) = 7, then f⁻¹(7) = 2. The graph of f⁻¹ is the mirror image of f in the line y = x, and the domain and range swap. Crucially, inverses only exist for one-to-one functions — the work classifying mappings from the last note pays off straight away.

📘 What you need to know

Definition: f⁻¹(x) reverses f(x). If f(a) = b, then f⁻¹(b) = a. Equivalently, f(f⁻¹(x)) = x and f⁻¹(f(x)) = x — one undoes the other.
Watch the notation: f⁻¹(x) is the inverse function, NOT the reciprocal 1/f(x). They are different things.
To find f⁻¹ algebraically: write y = f(x), swap x and y, then solve for y. The result is f⁻¹(x).
Graphically: reflect y = f(x) in the line y = x. A point (a, b) on f reflects to (b, a) on f⁻¹.
Domain & range swap: domain of f = range of f⁻¹; range of f = domain of f⁻¹.
When does f⁻¹ exist? Only when f is one-to-one. Many-to-one functions (like x²) have no inverse over all of ℝ.

The idea — what an inverse does

Think of f as a machine: input goes in, output comes out. f⁻¹ is the machine that runs in reverse — feed in an output and it gives back the original input. If f(x) = 2x, then f⁻¹(x) = x/2: doubling and halving cancel each other.

Inverse function — defining property if f(a) = b then f⁻¹(b) = a

f(f⁻¹(x)) = x & f⁻¹(f(x)) = x

Key warning: the −1 in f⁻¹(x) does NOT mean “to the power of −1”. It’s just notation for “the inverse”. The reciprocal 1f(x) = [f(x)]⁻¹ is a completely different thing. So for f(x) = 2x: the inverse is x/2, but the reciprocal is 12x.

Finding the inverse algebraically

The standard method has three steps: write the function as y = (something in x), swap the letters, then rearrange to make y the subject again. The result is f⁻¹(x).

Quick example: for f(x) = 3x + 5: write y = 3x + 5, swap to get x = 3y + 5, solve: y = (x − 5)/3. So f⁻¹(x) = (x − 5)/3.

If a question asks for f⁻¹(c) for a specific value c, you don’t need the full inverse formula — just solve f(x) = c for x. The value of x you get IS f⁻¹(c).

The graph: reflection in y = x

The graph of y = f⁻¹(x) is the mirror image of y = f(x) in the line y = x. So every point (a, b) on f becomes (b, a) on f⁻¹ — coordinates just swap. Any place where f crosses the line y = x is a fixed point: it sits on the mirror and so it’s also on f⁻¹.

The function (teal) and its inverse (orange) are mirror images across y = x. Picking any point on one and swapping its coordinates lands you on the other.

Domain & range swap

Because coordinates swap, the x-values and y-values trade places. So: domain of f⁻¹ = range of f, and range of f⁻¹ = domain of f. If f has domain 1 ≤ x ≤ 10 and range 5 ≤ f(x) ≤ 23, then f⁻¹ has domain 5 ≤ x ≤ 23 and range 1 ≤ f⁻¹(x) ≤ 10.

When does an inverse exist?

An inverse exists only when f is one-to-one. If two different inputs share one output, then “running the machine backwards” from that output gives two answers — which a function isn’t allowed to do. So f(x) = x² on all of ℝ has no inverse, because f(2) and f(−2) both equal 4. Graphically: a horizontal line on f that crosses the graph more than once means an inverse doesn’t exist (reflecting that horizontal line becomes a vertical line on f⁻¹, which would fail the vertical line test).

🧭 Recipe — finding an inverse function

Write y = f(x): set up the equation using y in place of f(x).
Swap the letters: replace every x with y and every y with x. This is the reflection-in-y = x idea, in algebra form.
Solve for y: rearrange the new equation to make y the subject again. Standard algebra — expand brackets, move terms, divide.
Relabel: replace y with f⁻¹(x). That’s your inverse function.
State the domain of f⁻¹ if asked — it’s the same set of numbers as the range of f. (And verify by checking one round-trip: f⁻¹(f(a)) should equal a.)

Worked examples

WE 1

Find the inverse algebraically

The function f is given by f(x) = 4x − 7. Find f⁻¹(x) and use it to find f⁻¹(13).

Step 1 — write y = f(x) y = 4x − 7 Step 2 — swap x and y x = 4y − 7 Step 3 — solve for y 4y = x + 7 y = (x + 7)/4 Step 4 — relabel and evaluate at 13 f ⁻¹(x) = (x + 7)/4 f ⁻¹(13) = (13 + 7)/4 = 20/4 = 5 f ⁻¹(x) = (x + 7)/4, f ⁻¹(13) = 5 round-trip check: f(5) = 4(5) − 7 = 13 ✓ — quick way to spot errors.

WE 2

Find f⁻¹(c) without the full formula

The function f is given by f(x) = x³ − 2. Find the value of f⁻¹(25).

Step 1 — recognise: f ⁻¹(25) is the x with f(x) = 25 solve: x³ − 2 = 25 Step 2 — rearrange x³ = 27 Step 3 — cube-root both sides x = ∛27 = 3 f ⁻¹(25) = 3 faster than finding the full inverse formula. Useful when only ONE value is asked for, not the whole function.

WE 3

Domain and range swap

The function g is defined by g(x) = 2x + 3 with domain 1 ≤ x ≤ 10. State the domain and range of g⁻¹.

Step 1 — find range of g first (linear, increasing) g(1) = 5, g(10) = 23 range of g: 5 ≤ g(x) ≤ 23 Step 2 — swap domain and range domain of g ⁻¹ = range of g range of g ⁻¹ = domain of g domain of g ⁻¹: 5 ≤ x ≤ 23 · range: 1 ≤ g ⁻¹(x) ≤ 10 always work out the range of the ORIGINAL function first — that’s what becomes the domain of the inverse. Don’t try to do both in one step.

WE 4

Reflection in y = x — finding the inverse line

The line y = 3x − 6 is reflected in the line y = x. Find the equation of the reflected line.

Reflection in y = x ≡ finding the inverse start: y = 3x − 6 Swap x and y x = 3y − 6 Solve for y 3y = x + 6 y = (x + 6)/3 y = (x + 6)/3 or y = x/3 + 2 cross-check: original passes through (3, 3) (a fixed point on y = x), and so does the inverse: (3 + 6)/3 = 3 ✓. Fixed points always lie on both.

WE 5

Explain why an inverse does not exist

The function h is defined on all real numbers by h(x) = x² + 3. Explain why h does not have an inverse function.

Step 1 — check if h is one-to-one h(2) = 4 + 3 = 7 h(−2) = 4 + 3 = 7 two DIFFERENT inputs give the SAME output Step 2 — what would the inverse have to do? h ⁻¹(7) would need to give both 2 AND −2 but a function can only output ONE value per input h is many-to-one, so h ⁻¹ doesn’t exist vertical line test on the reflected parabola fails — a horizontal line on h crosses the curve twice, which becomes a vertical line crossing h ⁻¹ twice. Banned.

WE 6

Word problem — currency conversion

A bureau de change uses the function E(d) = 0.92d to convert d US dollars into euros.
(a) Find E(200). (b) Find E⁻¹(460) and explain what it represents.

(a) substitute d = 200 E(200) = 0.92 × 200 = 184 so $200 converts to €184 (b) find E ⁻¹(x) first y = 0.92x ⇒ swap: x = 0.92y y = x/0.92, so E ⁻¹(x) = x/0.92 Evaluate at 460 E ⁻¹(460) = 460/0.92 = 500 (a) €184 · (b) E ⁻¹(460) = $500 interpretation: €460 is worth $500 going back the other way. Inverse functions are the natural tool for “reverse the operation” word problems — unit conversions, decoding formulas, undoing rates.

💡 Top tips

Round-trip check: after finding f⁻¹, plug an easy x into f, then plug the result into f⁻¹. You should land back where you started.
For one value only, skip the formula: f⁻¹(c) is just the x that solves f(x) = c. Faster than rearranging the whole inverse.
Swap before you solve: trying to rearrange before swapping x and y usually leads to a tangle. Swap first, then solve cleanly.
Domain & range of f⁻¹: just swap. The new domain is the OLD range; the new range is the OLD domain. Find the original range first.
If the question asks “does an inverse exist?”, check whether f is one-to-one over the given domain. Many-to-one ⇒ no inverse.

⚠ Common mistakes

Confusing f⁻¹(x) with 1f(x): these are completely different. For f(x) = 2x, the inverse is x/2; the reciprocal is 1/(2x).
Forgetting to swap before solving: solving for x in y = 4x − 7 gives x = (y+7)/4 — you still need to rename to f⁻¹(x) = (x+7)/4.
Claiming inverses exist for many-to-one functions: x² on all of ℝ has no inverse. Always check the function is one-to-one first.
Mixing up domain and range of the inverse: they SWAP. New domain = old range, new range = old domain. Easy to flip them by accident.
Trying to “guess” an inverse: only works for the very simplest cases. For anything with brackets, fractions, or several terms, use the swap-and-solve method every time.

Up next: Graphing Functions & Their Key Features. With the language of functions and inverses in hand, the focus turns to actually drawing them — finding intercepts, turning points, asymptotes, and symmetry, all efficiently using your GDC. The skill is reading what a graph tells you, and translating GDC output into a clean labelled sketch.

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