Curriculum
Course: Pure Mathematics — Second Secondary
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Curriculum

Pure Mathematics — Second Secondary

Unit 1: Real Functions and Curve Sketching

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Unit 2: Exponents and Logarithms

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Unit 3: Limits

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Unit 4: Trigonometry

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Text lesson

The Inverse Function

Pure Mathematics — Second Secondary

The Inverse Function

Worked examples

Worked examples

Example 1: a linear function

Comparison aspectMain ideaKeywords
•
  • Find the inverse of f(x) = 2x − 6
  • then check the result
Findinversethencheck
Solutiony = 2x − 6. Swap: x = 2y − 6, so 2y = x + 6 and y = (x + 6) / 2. Therefore f−1(x) = (x + 6) / 2.SwapTherefore
Checkf(f−1(x)) = 2 × (x + 6) / 2 − 6 = x + 6 − 6 = x.

Example 2: a cubic function

  • Find the inverse of f(x) = x3 + 1.
  • Solution y = x3 + 1. Swap: x = y3 + 1, so y3 = x − 1 and y = ∛(x − 1). Therefore f−1(x) = ∛(x − 1), which is defined for all real x.

Example 3: restricting the domain

  • The function f(x) = x2 is not one-to-one on the real numbers, but it is one-to-one on [0, ∞[. Find its inverse on this domain.
  • Solution y = x2. Swap: x = y2, so y = ±√x. The range of f on [0, ∞[ is [0, ∞[, so we choose the positive root only. Therefore f−1(x) = √x, with domain [0, ∞[.

Example 4: a rational function

Comparison aspectMain ideaKeywords
•Find the inverse of f(x) = 3 / (x − 1)Findinverse
Solutiony = 3 / (x − 1). Swap: x = 3 / (y − 1), so y − 1 = 3 / x and y = 1 + 3/x. Therefore f−1(x) = 1 + 3/x, with domain all real numbers except 0. This domain equals the range of f, because f never takes the value 0.SwapThereforedomainreal
Checkf(f−1(x)) = 3 / ((1 + 3/x) − 1) = 3 / (3/x) = x.

Example 5: a value of the inverse

Comparison aspectMain ideaKeywords
•If f(x) = 2x − 6, find f−1(4).find
Solutionf−1(4) = (4 + 6) / 2 = 5.
Check
  • f(5) = 2 × 5 − 6 = 4
  • which is correct
correct

Example 6: finding an inverse value of a quadratic

  • The function f(x) = x2 − 4 is defined on [0, ∞[. Find f−1(5).
  • Solution Solve f(x) = 5, that is x2 − 4 = 5, so x2 = 9 and x = 3 or x = −3. Since x must belong to [0, ∞[, we take x = 3. So f−1(5) = 3.
Check→f(3) = 9 − 4 = 5, which is correct.
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