The composition of functions, denoted as \(f \circ g\) (read as “f composed with g” or “f of g”), is defined as applying one function to the result of another. Mathematically:
In simpler terms, you first apply the function \(g\) to \(x\), and then you apply the function \(f\) to the result, \(g(x)\).
For the composition \(f(g(x))\) to be defined, the range of \(g\) must be a subset of the domain of \(f\).
This ensures that the output of \(g(x)\) is a valid input for \(f(x)\). If this condition is not met, the composite function is not defined for all \(x\) in the domain of \(g\).
The domain of the composite function \(f \circ g\) is the same as the domain of the inner function \(g\), provided the range condition is met.
Let \(f(x) = x^2\) and \(g(x) = x + 1\).
Find \((f \circ g)(x)\):
\$\((f \circ g)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1\)\$
Find \((g \circ f)(x)\):
\$\((g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1\)\$
Notice that in general, \(f \circ g \neq g \circ f\).
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\), \(f(x) = 2x - 1\) and \(g: \mathbb{R} \rightarrow \mathbb{R}\), \(g(x) = 3x^2\).
Find \(f \circ g\):
\$\(f(g(x)) = f(3x^2) = 2(3x^2) - 1 = 6x^2 - 1\)\$
\(dom(f \circ g) = \mathbb{R}\)
\(ran(f \circ g) = [-1, \infty)\)
Find \(g \circ f\):
\$\(g(f(x)) = g(2x - 1) = 3(2x - 1)^2 = 3(4x^2 - 4x + 1) = 12x^2 - 12x + 3\)\$
\(dom(g \circ f) = \mathbb{R}\)
\(ran(g \circ f) = [0, \infty)\)
| Function Composition | Rule | Domain | Range | Condition |
|---|---|---|---|---|
| \((f \circ g)(x) = f(g(x))\) | Evaluate \(f\) at \(g(x)\) | \(dom(g)\) such that \(g(x) \in dom(f)\) | Depends on \(f\) and \(g\) | \(r_g \subseteq d_f\) |
| \((g \circ f)(x) = g(f(x))\) | Evaluate \(g\) at \(f(x)\) | \(dom(f)\) such that \(f(x) \in dom(g)\) | Depends on \(f\) and \(g\) | \(r_f \subseteq d_g\) |
Free exam-style questions on Composition of Functions with instant AI feedback.
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