Compute (f∘g)(x), (g∘f)(x), (f∘f)(x), and (g∘g)(x) with full substitution steps. Evaluate at a specific x-value and explore domain restrictions — see exactly why order matters.
^ for powers, sqrt() for radicals, abs() for absolute value, * for multiplication. Coefficients like 2x are recognized automatically.
(f∘g)(x) = f(g(x)) ≠ g(f(x)) = (g∘f)(x) in general
What "apply g first, then f" means: In (f∘g)(x) you input x into g to get g(x), then feed that output as the input to f. The final result is f(g(x)).
Why order matters: Try f(x) = x² and g(x) = x + 1. Then (f∘g)(x) = (x+1)² = x²+2x+1, but (g∘f)(x) = x²+1 — completely different polynomials.
When they are equal: (f∘g) = (g∘f) only in special cases, notably when f and g are inverses of each other — because then each undoes the other.
Dom(f∘g) = { x ∈ Dom(g) : g(x) ∈ Dom(f) }
Two-step check: x must (1) be in the domain of g so that g(x) is defined, and (2) g(x) must land inside the domain of f so that f can accept it. Both conditions must hold simultaneously.
Example — radical: If f(x) = √x and g(x) = x − 4, then (f∘g)(x) = √(x−4) requires x − 4 ≥ 0, so the domain of f∘g is x ≥ 4, even though both f and g alone have different domains.
Example — rational: If f(x) = 1/x and g(x) = x + 1, then (f∘g)(x) = 1/(x+1) excludes x = −1, even though g(x) is defined everywhere.
One-on-one Precalculus tutoring walks through substitution, domain analysis, and inverse verification on your exact homework — building the fluency that makes compositions feel automatic.