Composition and inverse: machines in a row, and the undo button

The problem

Put on socks, then shoes. Now try shoes, then socks. Same two actions, very different result. Functions are the same: doing f then g is usually not the same as g then f.

And sometimes you want to take the shoes off again: to undo a function and get back your input. That undo machine is the inverse function.

The composition of f and g is (gof)(x) = g(f(x)): apply f first, then g. In general gf(x) ≠ fg(x). The inverse f⁻¹ of a one-one function reverses it: f⁻¹(y) = x exactly when y = f(x).

Key terms

Composition gf

gf(x) = g(f(x)): the output of f becomes the input of g. With f(x) = 2x + 3 and g(x) = x², gf(x) = g(2x + 3) = (2x + 3)².

Order matters

fg means g first, then f; gf means f first, then g. We also write ff as f² and fff as f³.

Inverse function f⁻¹

For a one-one function f from X onto Y, f⁻¹ goes from Y back to X. Then f⁻¹(f(x)) = x and f(f⁻¹(y)) = y: both compositions are identity maps.

Algebraic method

Write y = f(x), solve for x in terms of y, then rename y as x. Domain and range swap: domain f⁻¹ = range f, and range f⁻¹ = domain f.

Short questions with model answers

1. Q1. Let f(x) = 2x + 1 and g(x) = x² − 1. Find fg(x) and gf(x).

   • fg(x) = f(x² − 1) = 2(x² − 1) + 1 = 2x² − 1
   • gf(x) = g(2x + 1) = (2x + 1)² − 1 = 4x² + 4x

   fg(x) = 2x² − 1 · gf(x) = 4x² + 4x

2. Q2. Find f⁻¹(x) for f(x) = 2x + 1 and verify it.

   • y = 2x + 1 ⇒ x = ½(y − 1)
   • f⁻¹(x) = ½(x − 1)
   • f(f⁻¹(x)) = 2 · ½(x − 1) + 1 = x

   f⁻¹(x) = ½(x − 1)

3. Q3. Without finding the inverse, state the domain and range of f⁻¹ for f(x) = 2 + √(x − 1).

   • Domain f = [1, +∞)
   • Range f = [2, +∞)
   • Domain f⁻¹ = [2, +∞), range f⁻¹ = [1, +∞)

   Domain f⁻¹ = [2, +∞) · Range f⁻¹ = [1, +∞)

Common mistakes

• ✗ Reading gf(x) left to right as “g first, then f”.

  ✓ gf(x) = g(f(x)): the function next to x acts first. So f acts first, then g.

• ✗ Writing f⁻¹(x) = 1/f(x).

  ✓ f⁻¹ is the undo function, not the reciprocal. For f(x) = 2x + 1, f⁻¹(x) = ½(x − 1), not 1/(2x + 1).

• ✗ Giving f⁻¹ the same domain as f.

  ✓ They swap: the domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.

MCQs

1. 1. If f(x) = 2x + 1, then f²(x) = f(f(x)) is:

   1. (a) 4x² + 4x + 1
   2. (b) 4x + 3
   3. (c) 4x + 2
   4. (d) 2x² + 1
   Show answer

   (b) f(2x + 1) = 2(2x + 1) + 1 = 4x + 3. f² means f twice, not f squared.

2. 2. If g(x) = x² − 1, then g²(x) = g(g(x)) is:

   1. (a) x⁴ − 1
   2. (b) x⁴ − 2x²
   3. (c) x⁴ − 2x² + 2
   4. (d) (x² − 1)²
   Show answer

   (b) (x² − 1)² − 1 = x⁴ − 2x² + 1 − 1 = x⁴ − 2x².

3. 3. For f(x) = −2x + 8, f⁻¹(−1) is:

   1. (a) 10
   2. (b) 4.5
   3. (c) −4.5
   4. (d) 3.5
   Show answer

   (b) f⁻¹(x) = (8 − x)/2, so f⁻¹(−1) = 9/2 = 4.5. Check: f(4.5) = −9 + 8 = −1.

4. 4. For f(x) = 2 + √(x − 1), the domain of f⁻¹ is:

   1. (a) [1, +∞)
   2. (b) [2, +∞)
   3. (c) (−∞, 2]
   4. (d) all real numbers
   Show answer

   (b) Domain f⁻¹ = range f = [2, +∞).

5. 5. f⁻¹(f(x)) equals:

   1. (a) f(x)
   2. (b) 1
   3. (c) x
   4. (d) f⁻¹(x)
   Show answer

   (c) Undoing f returns the original input: f⁻¹∘f is the identity map.

Quick revision

• gf(x) = g(f(x)): f first, then g. In general gf(x) ≠ fg(x); f² means f(f(x)).
• A one-one onto function has an inverse; f⁻¹(f(x)) = x and f(f⁻¹(y)) = y.
• To find f⁻¹: set y = f(x), solve for x, rename. Domain and range swap.