What is an Inverse?
The big idea — switch x and y
Switch every (x, y) → (y, x)
An inverse relation undoes a given relation. Found by switching every x and y value — in a table, a graph, or an equation.
Notation: f⁻¹(x)
The inverse function of f(x) is written as f⁻¹(x). The −1 is NOT an exponent — it means "inverse of".
f⁻¹(x) ≠ 1/f(x)
Graph: reflect over y = x
The graph of f⁻¹ is the reflection of f over the line y = x. Points (a,b) on f become (b,a) on f⁻¹.
Domain ↔ Range swap
The domain of f is the range of f⁻¹, and the range of f is the domain of f⁻¹.
Inverses from Tables
Example 1 — just swap the rows
📌 Example 1 — Find the inverse relation of the table. Is it a function?
Original f
| x | 1 | 3 | 4 | 6 |
|---|---|---|---|---|
| y | −1 | 2 | 0 | 2 |
Inverse (swap rows)
| x | −1 | 2 | 0 | 2 |
|---|---|---|---|---|
| y | 1 | 3 | 4 | 6 |
✓
Original is a function — each x has exactly one y.✗
Inverse is NOT a function — x = 2 maps to both 3 and 6.📌 Part II Table — f and its inverse g = f⁻¹. Evaluate using both rows.
| x | −3 | −2 | 0 | 1 | 4 | 6 |
|---|---|---|---|---|---|---|
| f(x) | 6 | 3 | 1 | −1 | −3 | −7 |
| x | 6 | 3 | 1 | −1 | −3 | −7 |
| g(x)=f⁻¹(x) | −3 | −2 | 0 | 1 | 4 | 6 |
a
f(f(0)) = f(1) = −1 (f(0)=1 from table, then f(1)=−1)b
g(−3) = f⁻¹(−3) = 4 (read from inverse rows)c
g(6) = f⁻¹(6) = −3d
g(g(−1)) = g(1) = 0 (g(−1)=1, then g(1)=0)e
(f⁻¹ ∘ f)(−2) = f⁻¹(f(−2)) = f⁻¹(3) = g(3) = −2f
f⁻¹(−3) = g(−3) = 4Inverses from Graphs
Reflect over y = x — Examples 2, 3
Steps to Sketch an Inverse Graph (Linear Pieces)
1
List the key points (corners, endpoints) in a table.2
Create a new table for the inverse by switching x and y.3
Plot the new points and connect them with the same shape as the original.Steps to Sketch an Inverse Graph (Nonlinear Pieces)
1
Sketch the line y = x on the graph.2
Points already on the line y = x stay fixed.3
For other points, switch x and y (or reflect perpendicularly over y = x).4
Connect the new points following the same pattern as the original.📌 Example 2 — Key points: (−2,−1),(−1,2),(2,1),(4,3). Find inverse points.
Key points
| x | −2 | −1 | 2 | 4 |
|---|---|---|---|---|
| y | −1 | 2 | 1 | 3 |
Inverse points
| x | −1 | 2 | 1 | 3 |
|---|---|---|---|---|
| y | −2 | −1 | 2 | 4 |
Reading f⁻¹ from a graph
To find f⁻¹(a) from a graph: find the x-value where the original function equals a. That x-value IS f⁻¹(a).
Example: f⁻¹(6) = 11 means "f = 6 when x = 11", so the x-value where f equals 6 is 11.
📌 Part II Ex 3 — Graph of f on [−2, 8]. f⁻¹ questions.
a
Maximum value of f⁻¹(x)? The max of f⁻¹ is the max x-value of f = 8.b
f⁻¹(3) = 2 (x-value where f = 3); f⁻¹(1) = 3 (x-value where f = 1).c
Domain of f⁻¹ = Range of f = [−3, 7].Inverses from Equations
Examples 3 & 4 — algebraic method
Steps to Find f⁻¹(x) Algebraically
1
Replace f(x) with y.2
Switch x and y.3
For rational functions: multiply both sides by the denominator to clear the fraction.4
Move all terms with y to one side, everything else to the other.5
Factor out y from the left side.6
Divide both sides to isolate y. Rewrite as f⁻¹(x).📌 Example 3 — Find the inverse of f(x) = 3x − 7.
f(x) = 3x − 7
1
Write: y = 3x − 72
Switch x and y: x = 3y − 73
Solve for y: x + 7 = 3y → y = (x + 7)/3f⁻¹(x) = (x + 7) / 3
📌 Example 4a — Find the inverse of f(x) = (x − 2) / (x + 3).
f(x) = (x − 2) / (x + 3)
1
Switch x and y: x = (y − 2)/(y + 3)2
Multiply both sides by (y + 3): x(y + 3) = y − 2 → xy + 3x = y − 23
Move y terms left: xy − y = −3x − 24
Factor: y(x − 1) = −(3x + 2)5
Divide: y = −(3x + 2)/(x − 1)f⁻¹(x) = −(3x + 2) / (x − 1)
📌 Example 4b — Find the inverse of f(x) = (2x + 1) / (x − 3).
f(x) = (2x + 1) / (x − 3)
1
Switch x and y: x = (2y + 1)/(y − 3)2
Multiply: x(y − 3) = 2y + 1 → xy − 3x = 2y + 13
Move y terms left: xy − 2y = 3x + 14
Factor: y(x − 2) = 3x + 15
Divide: y = (3x + 1)/(x − 2)f⁻¹(x) = (3x + 1) / (x − 2)
Verifying Inverses
Examples 6, 7, 8 — must show BOTH compositions equal x
The Verification Theorem
f and g are inverses if and only if both of these hold:
f(g(x)) = x AND g(f(x)) = x
You must show both compositions. One alone is not enough for full credit.
📌 Example 6 — Show f(x) = 2x − 3 and g(x) = ½x + 3/2 are inverses.
1
f(g(x)) = 2(½x + 3/2) − 3 = (x + 3) − 3 = x ✓2
g(f(x)) = ½(2x − 3) + 3/2 = (x − 3/2) + 3/2 = x ✓Both compositions equal x → they ARE inverses ✓
📌 Example 7 — Show h(x) = x² + 10 and k(x) = √(x − 10) are inverses for x ≥ 10.
1
h(k(x)) = (√(x−10))² + 10 = (x−10) + 10 = x ✓ (since x−10 ≥ 0)2
k(h(x)) = √(x²+10−10) = √x² = |x| = x ✓ (since x ≥ 10 > 0)Both equal x (with restricted domain x ≥ 10) → inverses ✓
📌 Example 8 — Show f(x) = 6/(x−4) and g(x) = 6/x + 4 are inverses (x ≠ 0, x ≠ 4).
1
f(g(x)) = 6/(6/x + 4 − 4) = 6/(6/x) = 6 · (x/6) = x ✓2
g(f(x)) = 6/[6/(x−4)] + 4 = (x−4) + 4 = x ✓Both compositions equal x → inverses ✓
Key Properties of Inverses
From Part II — four facts to memorise
Property 1 — Notation
g(x) = f⁻¹(x)
If g is the inverse of f, then g(x) = f⁻¹(x).
Property 2 — Points swap
(a, b) on f → (b, a) on f⁻¹
Every (x, y) pair gets its coordinates swapped.
Property 3 — Domain/Range swap
Domain of f = Range of f⁻¹
Range of f = Domain of f⁻¹. Everything switches.
Property 4 — Invertibility
Must be strictly monotone
A continuous function has an inverse function only if it is strictly increasing OR strictly decreasing (passes the horizontal line test).
📌 Part II Ex 4 — g graph on [−3,10], h table. Find combined compositions.
| x | −5 | −1 | 0 | 2 | 5 | 6 |
|---|---|---|---|---|---|---|
| h(x) | −3 | 0 | 3 | 5 | 8 | 10 |
| x | −3 | 0 | 3 | 5 | 8 | 10 |
| h⁻¹(x) | −5 | −1 | 0 | 2 | 5 | 6 |
a
g(h(6)) = g(10) = −1 (h(6)=10 from table; g(10)=−1 from graph)b
g⁻¹(h(0)) = g⁻¹(3) = 4 (h(0)=3; find x where g=3 on graph → x=4)c
h⁻¹(g(8)) = h⁻¹(0) = −1 (g(8)=0 from graph; h⁻¹(0)=−1 from inverse table)d
h⁻¹(g⁻¹(−1)) = h⁻¹(10) = 6 (g⁻¹(−1)=10 since g(10)=−1; h⁻¹(10)=6)Quick Reference
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🔑 Algebraic Steps
Step 1: replace f(x) with y
Step 2: switch x and y
Step 3: solve for y
Step 4: write as f⁻¹(x) = ...
⚠️ Common Mistakes
❌ f⁻¹(x) = 1/f(x)
f⁻¹ means "inverse function" — NOT the reciprocal. 1/f(x) is completely different.
❌ Only showing one composition
To verify inverses you MUST show both f(g(x))=x AND g(f(x))=x.
❌ Every relation has an inverse function
The inverse relation always exists. But it's only a function if the original passes the horizontal line test (strictly monotone).