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Grade 12 · Functions and Graphs
Inverse Functions (Grade 12)
MEMORANDUM
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Mark
  / 20

Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.

  1. 1
    Determine the equation of h−1, the inverse of h(x) = 3x − 6, in the form y = …
    (2)
    x = 3y − 6 → y = 1/3x + 2✓✓ (2)
  2. 2
    Determine the inverse of f(x) = 2x².
    (2)
    x = 2y² → y = ±√(x/2)✓✓ (2)
  3. 3
    Is the inverse of f(x) = 2x² a function? Give a reason.
    (2)
    No — because of the ±, one x-value gives two y-values, so it fails the vertical line test.✓✓ (2)
  4. 4
    Restrict the domain of f(x) = 2x² so that f−1 is a function.
    (1)
    x ≥ 0 (or x ≤ 0)✓ (1)
  5. 5
    Determine the equation of k−1 if k(x) = 2x.
    (2)
    y = log₂ x✓✓ (2)
  6. 6
    Write down the domain and range of k−1 if k(x) = 2x.
    (2)
    Domain x > 0; range y ∈ ℝ✓✓ (2)
  7. 7
    The graph of f−1 is the reflection of f in which line?
    (1)
    y = x✓ (1)
  8. 8
    The point (2 ; 7) lies on f. Write down a point on f−1.
    (1)
    (7 ; 2)✓ (1)
  9. 9
    State the definition of a function.
    (2)
    A rule in which every input value has exactly ONE output value ✓✓(2)
  10. 10
    Explain, using the definition, why the inverse of f(x) = 3x² is not a function unless the domain is restricted.
    (3)
    f is many-to-one — f(−2) and f(2) both give 12 ✓; reflecting swaps inputs and outputs, so one input would map to two outputs ✓; restricting to x ≥ 0 makes f one-to-one, so the inverse is a function ✓(3)
  11. 11
    Which line test tells you whether the INVERSE of a graph will be a function?
    (2)
    The horizontal line test applied to the ORIGINAL graph ✓; if a horizontal line cuts it more than once, the inverse is not a function ✓(2)
TOTAL: 20 marks
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