0580

Functions

Algebra and Sequences

Undoing the rule

  • The inverse of f, written f⁻¹, reverses what f does, so it turns each output back into the input that produced it
  • An output that came from a given input goes straight back to it, so f(a) = b guarantees f⁻¹(b) = a — a check that needs no algebra
  • Applying a function and then its inverse leaves the number untouched: ff⁻¹(x) = x and f⁻¹f(x) = x
  • That cancelling is the fastest route through a question asking to solve f⁻¹(x) = c, since applying f to both sides gives x = f(c) directly
  • The −1 here is not a power, so f⁻¹(x) is not 1 ÷ f(x)

Finding it algebraically

  • Replace f(x) with y
  • Swap every x and every y, which changes the letters without moving any term
  • Rearrange until y stands alone, using exactly the methods for changing the subject of a formula
  • Write the answer as f⁻¹(x) = …, with no y left anywhere
  • The first rearranging step earns a mark by itself, so show it
Exam tip

Give the range as the outputs and the domain as the inputs, and keep the letters straight — domains in x, ranges in f(x). Read gf(x) as f first, then g. On this specification a domain is nearly always a listed set of a few numbers rather than an inequality, so the range is a listed set too, and two of three correct outputs still earns a mark.

Worked example

Finding an inverse and checking it

f(x) = 4x − 7. Find f⁻¹(x).

Solution:

  • The inverse undoes the function, so it turns an output back into the input that produced it
  • Write the rule as an equation: y = 4x − 7
  • Swap the letters, because the inverse reverses the roles of input and output: x = 4y − 7
  • That swapped line scores a mark on its own, so put it down before rearranging
  • Now make y the subject. Add 7 to both sides: x + 7 = 4y
  • Divide both sides by 4, and the bracket matters because the whole of x + 7 is divided
  • y = (x + 7) ÷ 4, so f⁻¹(x) = (x + 7) ÷ 4
  • Check by running a value through both: f(3) = 12 − 7 = 5, and f⁻¹(5) = 12 ÷ 4 = 3, which is the input we started from
  • Note f⁻¹ means the inverse function, never 1 ÷ f(x)

Domain and range swap over

  • The inverse sends outputs back to inputs, so the two sets trade places
  • The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f
  • Rewrite the swapped set in the right letters: a domain in terms of x, a range in terms of f⁻¹(x)
  • With f(x) = 4x − 7 on 1 ⩽ x < 5 the range is −3 ⩽ f(x) < 13, so f⁻¹ has domain −3 ⩽ x < 13