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NotesMath AA HLTopic 2.14Inverses: domain restriction & self-inverse
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2.14.21 min read

Inverses: domain restriction & self-inverse

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Restrict the domain so an inverse exists
  • Self-inverse functions
One input per output, or no inverse: An inverse exists only when the function is one-to-one (passes the horizontal-line test). If two x's give the same y, there's no single way back.

Fix it by restricting the domain to a stretch where the function is one-to-one — e.g. f(x) = x² needs x ≥ 0.

IB-style question — restrict, then invert

f(x) = x² is not one-to-one on ℝ.

Restrict its domain so f⁻¹ exists, and find f⁻¹.

Step by step

  1. Restrict to a half where it's one-to-one, say x ≥ 0.
  2. Swap x and y and solve: y = x² ⇒ x = √y (positive root).

Final answer

Restrict to x ≥ 0; then f⁻¹(x) = √x.

Its own inverse: A self-inverse function is its own inverse: f(f(x)) = x, so f⁻¹ = f. Its graph is symmetric in the line y = x.

Classic examples: f(x) = 1/x and f(x) = a − x.

IB-style question — show self-inverse

Show that f(x) = (2x + 3)/(x − 2) is self-inverse.

Step by step

  1. Find f⁻¹: write y = (2x + 3)/(x − 2) and make x the subject.
  2. Solve for x.
  3. Swap to get f⁻¹(x) — it's the same rule as f.

Final answer

f⁻¹ = f, so f is self-inverse.

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Show that f(x) = 6 − x is self-inverse. [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

2.1.1Equations of lines
2.1.2Parallel lines
2.1.3Perpendicular lines
2.1.4Perpendicular bisector
View all Math AA HL topics

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