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Business Mathematics and Statistics · Ch 4 — Functions

Composite Functions and Inverse Functions

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Composite Functions and Inverse Functions

Composite function. Given two functions f:A→Bf: A \to B and g:B→Cg: B \to C, the composite function g∘f:A→Cg \circ f: A \to C (read 'g composed with f', or 'g of f') is defined by first applying ff and then applying gg to the result:

(g∘f)(x)=g(f(x))(g \circ f)(x) = g\big(f(x)\big)

For this composition to make sense, the codomain of ff must match (or be contained in) the domain of gg — the output of the first function must be a valid input to the second. Composition is used constantly in business modelling: for example, if qq is a function giving the quantity produced as a function of labour hours, and CC is a function giving total cost as a function of quantity produced, then C∘qC \circ q gives total cost directly as a function of labour hours.

Composition is generally not commutative — that is, f∘gf \circ g and g∘fg \circ f are, in general, two entirely different functions, even when both happen to be defined. It is always necessary to compute both separately rather than assuming they agree.

Inverse function. If f:A→Bf: A \to B is bijective (Section 4), then for every b∈Bb \in B there is exactly one a∈Aa \in A with f(a)=bf(a)=b. This lets us define a new function, the inverse function f−1:B→Af^{-1}: B \to A, by f−1(b)=af^{-1}(b) = a whenever f(a)=bf(a)=b. In effect, f−1f^{-1} simply reverses every ordered pair of ff: if (a,b)∈f(a,b) \in f, then (b,a)∈f−1(b,a) \in f^{-1}.

An inverse function undoes what the original function does, in the precise sense that

f−1(f(x))=xfor every x in the domain of ff^{-1}\big(f(x)\big) = x \quad \text{for every } x \text{ in the domain of } f

f(f−1(y))=yfor every y in the domain of f−1f\big(f^{-1}(y)\big) = y \quad \text{for every } y \text{ in the domain of } f^{-1} …

Definition 1Composite Function

For f:A→Bf: A \to B and g:B→Cg: B \to C, the composite (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x)) applies ff first and then gg. In general $f \c …

Definition 2Inverse Function

If f:A→Bf: A \to B is bijective, its inverse f−1:B→Af^{-1}: B \to A satisfies f−1(f(x))=xf^{-1}(f(x))=x and f(f−1(y))=yf(f^{-1}(y))=y. Found algebraically by writing y=f(x)y=f(x) and solv …