Math · College algebra · Worked example

Compose two functions in both orders

For f(x) = 2x + 1 and g(x) = x², find (f ∘ g)(x) and (g ∘ f)(x), and evaluate both at x = 3.

f (x) = 2 x + 1, g (x) = x^{2}

Find f(g(x))

Put g(x) = x² into f: double it and add 1.

f (g (x)) = 2 x^{2} + 1

Find g(f(x))

Put f(x) = 2x + 1 into g: square it.

g (f (x)) = {(2 x + 1)}^{2} = 4 x^{2} + 4 x + 1

\begin{array}{c} g (f (x)) = {(2 x + 1)}^{2} \\ = 4 x^{2} + 4 x + 1 \end{array}

Evaluate at 3

Inside first: g(3) = 9, then f(9) = 19. In the other order, f(3) = 7, then g(7) = 49.

\begin{array}{l} f (x) = 2 x + 1 \\ g (x) = x^{2} \\ g (3) = 9 \\ f (9) = 19 \\ f (3) = 7 \\ g (7) = 49 \end{array}

Compare the orders

19 and 49 differ, so the two composites are different functions: composition is not commutative.

Result

(f ∘ g)(x) = 2x² + 1 and (g ∘ f)(x) = (2x + 1)² = 4x² + 4x + 1. At x = 3 they give 19 and 49.

Your turn

For f(x) = √x and g(x) = x − 4, find f(g(x)) and its domain.

Show the answer and explanation

f(g(x)) = √(x − 4), for x ≥ 4.

Apply g first, then take the square root: √(x − 4). Every real x is allowed in g, but the square root needs x − 4 ≥ 0, so the domain is x ≥ 4.

f (g (x)) = \sqrt{x - 4}, x \geq 4

Keep exploring

In Graph, plot 2x² + 1 and (2x + 1)². The curves cross only where 2x² + 1 = (2x + 1)², at x = 0 and x = −2, so the two composites agree at just those inputs.

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