AHL 2.7 (HL)—Composite and inverse functions

The composite (f∘g)(x)=f(g(x)) applies g first and then f. Its domain is restricted wherever the inner output is outside the outer function's domain.

Keep the order visible. If f(x)=√x and g(x)=x−1, then (f∘g)(x)=√(x−1), so x≥1; reversing the order gives √x−1 with a different domain and meaning.

For f(x)=2x+1 and g(x)=x², (f∘g)(3)=2·9+1=19, while (g∘f)(3)=7²=49. The functions need not commute.

Composition is not multiplication. Check which function acts first and state the domain rather than assuming both orders are valid.

Inverse workflow: restrict the original domain until the function is one-to-one, write $y=f(x)$, swap $x$ and $y$, and solve for $y$. Then $(f\circ f^{-1})(x)=x$ on the range of $f$ and $(f^{-1}\circ f)(x)=x$ on the restricted domain. For $f(x)=(x-3)^2$ restricted to $x\ge3$, $f^{-1}(x)=3+\sqrt{x}$; choosing $x\le3$ would give the other branch $3-\sqrt{x}$.