AHL 2.7—Composite functions, finding inverse function incl domain restriction

Composition is not commutative in general: $(f\circ g)(x)\ne (g\circ f)(x)$. Always identify which function acts first.

Let $f(x)=x^2$ and $g(x)=x+1$. Find $(f\circ g)(x)$ and $(g\circ f)(x)$.

Finding $(f\circ g)(x)$:
$(f\circ g)(x)=f(g(x))=f(x+1)=(x+1{)}^2=x^2+2x+1$

Finding $(g\circ f)(x)$:
$(g\circ f)(x)=g(f(x))=g(x^2)=x^2+1$

Clearly $x^2+2x+1\ne x^2+1$ in general , composition is order-dependent.

Let $f(x)=\sqrt{x}$ and $g(x)=3x-2$. Evaluate $(f\circ g)(6)$.

Step 1: Apply $g$ first: $g(6)=3(6)-2=16$

Step 2: Apply $f$: $f(16)=\sqrt{16}=4$

So $(f\circ g)(6)=4$.

Alternatively, find the composite rule first: $(f\circ g)(x)=\sqrt{3x-2}$, then substitute $x=6$: $\sqrt{3(6)-2}=\sqrt{16}=4$. ✓

Composing three functions: Let $f(x)=x+1$, $g(x)=2x$, $h(x)=x^2$. Find $(f\circ g\circ h)(x)$.

Work from the innermost function outward:

Step 1: Apply $h$ first: $h(x)=x^2$

Step 2: Apply $g$: $g(h(x))=g(x^2)=2x^2$

Step 3: Apply $f$: $f(g(h(x)))=f(2x^2)=2x^2+1$

So $(f\circ g\circ h)(x)=2x^2+1$.

When evaluating a composite at a specific value, it is often faster to work numerically step by step rather than finding the general formula first.