Practice Maths

L42 — Inverse and Composite Functions

Key Terms

Composite function f ˆ g
"Apply g first, then f"; (f ˆ g)(x) = f(g(x)). The order matters — f ˆ g ≠ g ˆ f in general.
Inverse function f−1
The function that undoes f; found by swapping x and y in y = f(x) and solving for y.
Reflection in y = x
The graph of f−1 is the mirror image of the graph of f across the line y = x.
Domain restriction
To create an inverse for a many-to-one function, restrict the domain so the function becomes one-to-one.
f(f−1(x)) = x
The defining property of inverse functions — composing a function with its inverse returns the original input.
Domain of f−1
The domain of the inverse equals the range of the original function, and vice versa.

Composite functions

NotationMeaningRule
(f ∘ g)(x)"f of g of x"Substitute g(x) into f: f(g(x))
(g ∘ f)(x)"g of f of x"Substitute f(x) into g: g(f(x))
Order mattersf ∘ g ≠ g ∘ f in generalApply right-hand function first
Domain of f ∘ gx in domain of g where g(x) is in domain of fCheck both restrictions

Inverse functions

PropertyDetail
Notationf−1(x) — "the inverse of f"
EffectReverses the mapping: if f(a) = b, then f−1(b) = a
Finding itWrite y = f(x), swap x and y, solve for y
Cancellationf(f−1(x)) = x and f−1(f(x)) = x
GraphReflection of y = f(x) in the line y = x
ExistenceOnly one-to-one functions have an inverse that is also a function
x y 1 2 −1 −2 1 2 −1 y = x f(x)=2x+1 f⁻¹(x)=(x−1)/2 (0,1) (1,0)
f(x) = 2x + 1 and its inverse f−1(x) = (x−1)/2 are reflections in y = x.
Hot Tip: To find f−1(x): write y = f(x), swap x and y, then solve for y. Always check the domain: f−1 has the same domain as the range of f, and its range equals the domain of f.

Worked Example 1 — Composite functions

Given f(x) = 3x − 1 and g(x) = x² + 2, find: (a) f(g(x)); (b) g(f(x)); (c) f(g(2)).

(a) f(g(x)) = f(x²+2) = 3(x²+2) − 1 = 3x² + 5.

(b) g(f(x)) = g(3x−1) = (3x−1)² + 2 = 9x² − 6x + 1 + 2 = 9x² − 6x + 3.

(c) g(2) = 4+2 = 6. Then f(6) = 3(6)−1 = 17.

Worked Example 2 — Finding an inverse

Find the inverse of f(x) = (2x + 3)/5.

Step 1: Write y = (2x+3)/5.

Step 2: Swap x and y: x = (2y+3)/5.

Step 3: Solve for y: 5x = 2y+3 ⇒ 2y = 5x−3 ⇒ y = (5x−3)/2.

Inverse: f−1(x) = (5x − 3)/2.

Verify: f(f−1(x)) = (2×(5x−3)/2 + 3)/5 = (5x−3+3)/5 = 5x/5 = x. ✓

Worked Example 3 — Inverse of a restricted quadratic

Find the inverse of f(x) = x² − 4 with domain restricted to x ≥ 0.

y = x² − 4. Swap: x = y² − 4. Solve: y² = x+4 ⇒ y = ±√(x+4).

Since original domain is x ≥ 0 (so range is y ≥ −4), the inverse has domain x ≥ −4 and we take the positive root.

f−1(x) = √(x + 4), domain x ≥ −4.

See Answers ➔

  1. Composite functions — basic. Fluency

    Let f(x) = 2x + 1 and g(x) = x − 3. Find:

    • (a) f(g(x))
    • (b) g(f(x))
    • (c) f(f(x))
    • (d) f(g(4)) and g(f(4))

  2. Finding inverses — linear. Fluency

    Find the inverse of each function.

    • (a) f(x) = x + 7
    • (b) g(x) = 3x − 5
    • (c) h(x) = (x + 2)/4
    • (d) k(x) = (1 − 2x)/3

  3. Inverse verification. Fluency

    • (a) Show that f(x) = 5x − 3 and g(x) = (x+3)/5 are inverse functions by computing f(g(x)) and g(f(x)).
    • (b) If f(7) = 4, what is f−1(4)?
    • (c) If f−1(2) = −3, what is f(−3)?
    • (d) A function has f−1(x) = 2x − 6. Find f(x).

  4. Non-linear composites. Fluency

    Let p(x) = x² and q(x) = x + 4. Find:

    • (a) p(q(x))
    • (b) q(p(x))
    • (c) p(q(−2))
    • (d) Solve p(q(x)) = 9.

  5. Inverse function from a graph. Understanding

    The graph shows f(x) = √(x − 1) + 2 and its inverse f−1(x).

    x y 1 2 3 −1 1 2 −1 −2 y=x (1,2) f(x) (2,1) f⁻¹(x)
    Blue: f(x) = √(x−1) + 2 (domain x ≥ 1). Red: f−1(x) = (x−2)² + 1 (domain x ≥ 2).
    • (a) State the domain and range of f(x) = √(x − 1) + 2.
    • (b) Find f−1(x) algebraically. Verify using the graph.
    • (c) State the domain and range of f−1(x).
    • (d) Use the graphs to find all x where f(x) = f−1(x). What geometric feature do these points lie on?

  6. Composite functions — mixed. Understanding

    Let f(x) = √x and g(x) = x − 4. Find:

    • (a) f(g(x)) and its natural domain.
    • (b) g(f(x)) and its natural domain.
    • (c) Are f(g(x)) and g(f(x)) the same function? Explain.
    • (d) Find x such that f(g(x)) = 3.

  7. Inverse functions — non-linear. Understanding

    • (a) Find the inverse of f(x) = x³ + 1. State the domain and range of both f and f−1.
    • (b) f(x) = x² + 3 (no domain restriction). Explain why f does not have an inverse function. What restriction on the domain would allow an inverse?
    • (c) Find the inverse of f(x) = 2/x. What do you notice about f and f−1?
    • (d) If f(x) has domain [0, 5] and range [1, 11], state the domain and range of f−1.

  8. Composites with three functions. Understanding

    Let f(x) = x + 1, g(x) = 2x, h(x) = x².

    • (a) Find h(g(f(x))). Simplify.
    • (b) Find f(g(h(x))). Simplify.
    • (c) Evaluate h(g(f(3))).
    • (d) For what value of x is h(g(f(x))) = g(f(h(x)))?

  9. Applied composition and inverse. Problem Solving

    A factory converts raw material (kg) into products. Step 1: remove 5 kg of waste — modelled by f(x) = x − 5. Step 2: multiply the remainder by 3 (three units per kg) — modelled by g(x) = 3x.

    • (a) Find the composite g(f(x)) and interpret its meaning.
    • (b) If 50 kg of raw material enters, how many units are produced?
    • (c) Find the inverse of g(f(x)). What does the inverse represent?
    • (d) How much raw material is needed to produce 120 units?

  10. Decomposing functions. Problem Solving

    • (a) Express h(x) = (3x + 2)⁵ as a composition f(g(x)), stating f and g explicitly.
    • (b) Express h(x) = √(x² + 1) as a composition f(g(x)), stating f and g.
    • (c) A function has the property f(f(x)) = x for all x. Give an example, and prove it works using the definition of an inverse.
    • (d) Given f(x) = 2x + 1 and f(g(x)) = 6x − 3, find g(x).