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==Introduction== After learning about the definition of a function we learned about how to evaluate a function at a real number. Recallin..."

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==Introduction==

After learning about the definition of a function we learned about how to evaluate a function at a real number. Recalling how this is defined, if we want to evaluate a function f(x) at x = 5, we would replace all occurrences of x with 5, and simplify.
For composite functions, instead of replacing the independent variable, usually x, with a number, we replace it with a function.

==Definition and notation==

Given two functions, f and g, the composite function, denoted <math> f\circ g</math>, is a function where <math>(f\circ g) (x) = f(g(x))</math>.

Example:

Suppose <math>f(x) = \frac{1+x}{x - 3} \text{ and }g(x) = \sqrt{x}<math>

Then <math> ()f\circ g) (x) = \frac{1+\sqrt{x}}{\sqrt{x} - 3}</math>

==Domain==

The domain of a composite function <math>(f\circ g)(x)</math> is the collection of x-values in the domain of g such that g(x) is in the domain of f.

Example: Find the domain of <math> f \circ g \text{ if } f(x) = \frac{1}{x + 1} \text{ and } g(x) = \frac{1}{x + 3}</math>

We start by noting that the domain of g(x) is <math>(-\infty, 3) \cup (3, \infty)</math>. Now we want to know for what values of x is g(x) = -1.
So we solve: <math> -1 = \frac{1}{x + 3}</math>. Solving this equation we find that g(-4) = -1. So -4 must be removed from the domain of g to result in th domain of <math> f\circ g</math>.
To finish the problem: the domain of <math> f\circ g \text{ is }(-\infty, -4)\cup (-4, 3) \cup (3, \infty)</math>

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Composite Functions - Revision history

MathAdmin: Created page with "__TOC__ ==Introduction== After learning about the definition of a function we learned about how to evaluate a function at a real number. Recallin..."

MathAdmin: Created page with "
TOC
==Introduction== After learning about the definition of a function we learned about how to evaluate a function at a real number. Recallin..."