---

Warm-Up

Your weather app converts Celsius to Fahrenheit: F = 1.8C + 32. You're traveling, the forecast says 77°F, and you want Celsius. You don't need a new formula from the internet — you un-do the old one: subtract 32, divide by 1.8: C = (77 − 32)/1.8 = 25°C.

That's an inverse function: the machine run backwards. Every question about f⁻¹ is a question about f with input and output trading jobs. (And tomorrow's logarithms are nothing but this lesson applied to exponentials.)

---

Core Concept

When does an inverse exist?

f⁻¹ exists when f is one-to-one: every output comes from exactly one input. Graphically: every horizontal line crosses the graph at most once.

• Strictly increasing (or strictly decreasing) on its whole domain → invertible. (This is the practical test on the AP: monotonic ⇒ one-to-one.)
• Fails: parabolas, |x|, even functions generally, anything with a turnaround — two inputs share an output.

Domain restriction rescues non-invertible functions: f(x) = (x − 3)² isn't one-to-one, but restricted to x ≥ 3 (the right half, from the vertex on) it's strictly increasing, and the inverse is f⁻¹(x) = 3 + √x. You choose a piece on which the function is monotonic.

The three signature facts

If f(a) = b, then f⁻¹(b) = a. Everything else follows:

1. Reversed pairs: (a, b) on f ↔ (b, a) on f⁻¹.
2. Swapped domain and range: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.
3. Reflection: the graph of f⁻¹ is the graph of f reflected over the line y = x.

And the defining property, in composition language (Lesson 13's identity function):

f⁻¹(f(x)) = x  for all x in the domain of f
f(f⁻¹(x)) = x  for all x in the domain of f⁻¹

Composing a function with its inverse yields the identity — this is the verification test for any inverse you compute.

Finding f⁻¹ analytically

The input/output swap, as a procedure:

1. Write y = f(x).
2. Solve for x in terms of y (undo operations in reverse order).
3. The result expresses f⁻¹: swap letters at the end if you want f⁻¹ written as a function of x.

f(x) = x³ + 1:   y = x³ + 1  →  x³ = y − 1  →  x = (y − 1)^(1/3)
f⁻¹(x) = (x − 1)^(1/3)

Verify: f(f⁻¹(x)) = ((x−1)^(1/3))³ + 1 = x − 1 + 1 = x ✓

⚠️ Notation trap: f⁻¹(x) is NOT 1/f(x). The −1 is "inverse function," not an exponent. The reciprocal of f is written (f(x))⁻¹.

Self-inverse curiosities

Some functions undo themselves: f(x) = 1/x satisfies f(f(x)) = x, as does f(x) = −x + b (any b) — their graphs are symmetric over y = x already. Recognizing these saves time on "find the inverse" MC items.

[GRAPH: The line y = x dashed diagonally. f(x) = x³ + 1 rising through (0, 1) and (1, 2); its inverse (x − 1)^(1/3) rising through (1, 0) and (2, 1) — each the mirror of the other across the dashed line. Points (0,1) and (1,0) marked as a reflected pair.]

---

Worked Examples

Example 1 (easy) — Linear inverse 🚫 No-Calc

Problem: f(x) = 2x − 6. Find f⁻¹(x) and verify by composition.

Solution: y = 2x − 6 → x = (y + 6)/2 → f⁻¹(x) = (x + 6)/2. Verify: f(f⁻¹(x)) = 2·(x + 6)/2 − 6 = x + 6 − 6 = x ✓

Interpretation: f does "double, then subtract 6"; f⁻¹ does "add 6, then halve" — inverse operations in reverse order, like taking off shoes and socks.

Example 2 (medium) — Inverse values without formulas 🚫 No-Calc

Problem: f is one-to-one with f(3) = 8 and f(5) = 2. Find f⁻¹(8), f⁻¹(2), and (f⁻¹∘f)(3).

Solution: f⁻¹(8) = 3 and f⁻¹(2) = 5 (reversed pairs). (f⁻¹∘f)(3) = 3 automatically — composing with the inverse is the identity.

Interpretation: No formula needed; inverse questions about specific values are pair-reversal questions. If you're solving equations, you're working too hard.

Example 3 (medium) — Rational inverse with domain bookkeeping 🚫 No-Calc

Problem: f(x) = (2x + 1)/(x − 3). Find f⁻¹(x), and state the domain of f⁻¹.

Solution:

y(x − 3) = 2x + 1  →  xy − 3y = 2x + 1  →  x(y − 2) = 3y + 1  →  x = (3y + 1)/(y − 2)
f⁻¹(x) = (3x + 1)/(x − 2)

Domain of f⁻¹ = range of f. f has horizontal asymptote y = 2 (equal degrees, ratio 2/1) and takes every other value → range of f is y ≠ 2 → domain of f⁻¹: x ≠ 2 (visible in the formula's denominator ✓).

Interpretation: Collect x-terms on one side, factor x out — the standard move for rational inverses. The asymptote of f reappears as the excluded input of f⁻¹: swapped roles, as always.

Example 4 (AP-style) — Restricting to invert 🚫 No-Calc

Problem: g(x) = x² − 4x + 7. Explain why g is not invertible on its natural domain, restrict its domain to make it invertible, and find the inverse on your restriction.

Solution: g(x) = (x − 2)² + 3: a parabola with vertex (2, 3) — fails the horizontal line test (g(1) = g(3) = 4). Restrict to x ≥ 2 (g strictly increasing there). Invert: y = (x − 2)² + 3 → (x − 2)² = y − 3 → x − 2 = √(y − 3) (nonnegative root, since x ≥ 2) → x = 2 + √(y − 3). g⁻¹(x) = 2 + √(x − 3), with domain x ≥ 3 (the range of restricted g).

Interpretation: The restriction dictates the sign of the root. Choosing x ≤ 2 instead would give 2 − √(x − 3). Either restriction is legal; consistency between restriction and root sign is what's graded.

---

Common Mistakes

1. f⁻¹(x) = 1/f(x). The notation clash claims victims every year. Inverse function ≠ reciprocal. Fix: read f⁻¹ as "f-undone."
2. Claiming every function has an inverse. Parabolas and friends don't (two inputs, one output). Fix: horizontal line test, or check monotonicity.
3. Forgetting the domain/range swap. The inverse's domain isn't "all x the formula allows" — it's the range of the original. Example 3's x ≠ 2 came from f's asymptote, not from staring at f⁻¹'s formula (though they agree when you've done it right).
4. Wrong root sign after restricting. x ≥ 2 forces the + root; x ≤ 2 the − root. Fix: check one point (g(4) = 7 → g⁻¹(7) should be 4: 2 + √4 = 4 ✓).
5. Verifying only one composition. For full rigor check both f(f⁻¹(x)) = x and f⁻¹(f(x)) = x; on the AP, one composition plus correct domains is typically accepted — but write the composition, don't skip verification.

---

Practice Problems

12. (FRQ-style) 🚫 h(x) = 5·2ˣ. (i) Explain why h is invertible (use monotonicity). (ii) Find h⁻¹(20) and h⁻¹(5) exactly. (iii) State the domain and range of h⁻¹, and explain how each follows from h.

---

FRQ Practice — Task Model: Symbolic Manipulations (FRQ 3 style) 🚫 No-Calc

Let f(x) = (2x + 1)/(x − 3).

(a) (i) Find f⁻¹(x). (ii) Verify your answer by computing f(f⁻¹(x)) and simplifying.

(b) (i) State the domain and range of f. (ii) State the domain and range of f⁻¹, and explain the relationship.

(c) Solve f(x) = 4 two ways: directly, and by evaluating your f⁻¹ at 4. Confirm the answers agree.

Model Response & Rubric (6 points)

(a) [2 pts] (i) [1 pt] y(x − 3) = 2x + 1 → x(y − 2) = 3y + 1 → f⁻¹(x) = (3x + 1)/(x − 2). (ii) [1 pt] f(f⁻¹(x)) = [2·(3x+1)/(x−2) + 1] / [(3x+1)/(x−2) − 3] = [(6x + 2 + x − 2)/(x−2)] / [(3x + 1 − 3x + 6)/(x−2)] = (7x)/(7) = x ✓

(b) [2 pts] (i) [1 pt] Domain of f: x ≠ 3. Range: y ≠ 2 (horizontal asymptote y = 2 from equal degrees, ratio 2; a rational function of this form attains every value except its asymptote). (ii) [1 pt] f⁻¹ has domain x ≠ 2 and range y ≠ 3 — exactly f's range and domain, swapped, because inverses reverse the roles of input and output.

(c) [2 pts] Directly: (2x + 1)/(x − 3) = 4 → 2x + 1 = 4x − 12 → 13 = 2x → x = 13/2 [1 pt]. Via inverse: f⁻¹(4) = (12 + 1)/(4 − 2) = 13/2 ✓ agree [1 pt]. (Solving f(x) = c IS evaluating f⁻¹(c) — the same act in two notations.)

---

Show answer key & explanations

(g) Answer Key

1. (A). Undo in reverse: add 6, halve: (x + 6)/2. (B) halves before adding; (D) is the reciprocal trap.

2. (B). Reversed pair: (3, 8) on f → (8, 3) on f⁻¹ → f⁻¹(8) = 3. (A) is the reciprocal trap again.

3. (C). Strictly monotonic ⇒ one-to-one ⇒ invertible. (A)/(D) fail the horizontal line test; (B) catastrophically so.

4. (D). Swapping (a, b) ↔ (b, a) is reflection across y = x.

5. (B). From the vertex (x = 3) rightward, f is strictly increasing. (A) still contains the vertex and both sides near it (f(0) = 9 = f(6) — two inputs, one output on x ≥ 0).

6. (A). Which input gives 7? f(2) = 7 → f⁻¹(7) = 2.

7. (C). Function-then-inverse is the identity: x. That's the defining equation.

8. (D). Swap: domain [−1, 5] (old range), range [2, 9] (old domain). No sign changes — (B)/(C) invent reflections that aren't part of the swap.

9. (A). Undo "+1" then "cube": subtract 1, cube-root: (x − 1)^(1/3). (B) undoes in the wrong order; (C) reciprocal trap.

10. (C). f(f(x)) = 1/(1/x) = x ✓. (A) composes to 4x; (D) to x + 2.

11. (B). Two inputs sharing an output makes reversal ambiguous — the horizontal line test in words. (A) is irrelevant (decreasing functions invert fine).

12. (FRQ-style, 6 points) (i) [2 pts] h is an exponential with base 2 > 1 and positive coefficient: strictly increasing on all of ℝ, hence one-to-one and invertible. (ii) [2 pts] h⁻¹(20): solve 5·2ˣ = 20 → 2ˣ = 4 → x = 2. h⁻¹(5): 5·2ˣ = 5 → 2ˣ = 1 → x = 0. (iii) [2 pts] h has domain ℝ and range (0, ∞) → h⁻¹ has domain (0, ∞) and range ℝ, by the domain/range swap. (h's asymptote at 0 means h never outputs 0 or below — so h⁻¹ never receives such inputs.)

---

🎯 Exam tip: Problem 12 is tomorrow's lesson in disguise: h⁻¹ is a logarithm. Every log fact you're about to learn — domain (0, ∞), range ℝ, the reflection of the exponential — is this lesson's swap rules applied to a·bˣ. Nail the swap now and logs arrive pre-explained.