MathematicsExam code: 0580

Functions

Algebra and Sequences

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)

Worked example

Solving h⁻¹(x) = c without finding the inverse

h(x) = 3ˣ. Find x when h⁻¹(x) = 4, and find x when h⁻¹(x) = −2.

Solution:

h⁻¹(x) = 4 says the inverse sends x to 4, so h sends 4 back to x
So x = h(4): write this line first, because it earns a method mark on its own
x = 3⁴ = 81
In the same way, h⁻¹(x) = −2 gives x = h(−2) = 3⁻² = 1 ÷ 3² = 1/9
Check: h(4) = 81, so h⁻¹ takes 81 back to 4, as the question said
There is no need to find h⁻¹(x), and for 3ˣ you could not: swapping the letters gives x = 3ʸ, which no rearranging in this course can solve for y

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

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)

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

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