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Tuesday, November 27, 2007

Compositing functions

Composite functions are what you get when you take the output of one function and use it for the input of the next one. In this discussion, we will discuss the composition of functions which are R1 R1, i.e. a real number as an input and a real number as an output. The notation for this is (fg)(x)=f(g(x)), where the output of g(x) becomes the input of f(x) and is described as (fg)(x). As a real example, let's use f(x)=x2+2*x-2, and g(x)=3*x+2. By replacing all of the occurrances of x in f(x) by the formula for g(x) we can find the formula in x for the composite function. So:
(fg)(x) = f(g(x)) = f(3*x+2)
(fg)(x) = (3*x+2)2+2*(3*x+2)-2
(fg)(x) = (9*x2+12*x+4)+(6*x+4)-2
(fg)(x) = 9*x2+18*x+6

We can also find (gf)(x):
(gf)(x) = g(f(x)) = g(x2+2*x-2)
(gf)(x) = 3*(x2+2*x-2)+2
(gf)(x) = 3*x2+6*x-4
That's basic composite functions. Compositing functions can do weird things to the domain and range, though, because your doing mappings onto mappings (i.e. mapping one domain to another domain to a range). One important point is that the composite of a function with its inverse yields an identity function. That is f(f-1(x))=f-1(f(x))=x. What's happening is that your mapping the original function to its range and then straight back into the domain.


Composite Functions

Functions can be combined to give a composite function (sometimes called a function of a function).

Example 1

Two functions are given f(x) = x + 1 and g(x) = 3x. Illustrate the composite function derived by operating with f first on x and then with g on the result.

Starting with x = 1, f(1) = 2. Now operating with g we have g[f(1))] = g(2) = 6.

The composite function is produced from g[f(x)] which is written gf(x).
Note that this is written as gf but f is the first function to operate, followed by g. (It does NOT mean g x f.
Instead of calculating the results one by one, we can produce a formula for the composite function gf.

So gf: x 3x + 3. gf(x) is not the same as fg(x). The order of a composite function is important.

Example 2

A function f is defined by f:x x2 + 5x - 5 for x>0. Find the value of x which is unchanged by the mapping.

Since the image of x is x,

f(x) = x
i.e. x2 + 5x - 5 = x
x2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -1 or 5
Since -5 is not in the domain, the value is 1.

Example 3

A function f is defined by f:x2x/(x - 1), x1.

(a) Obtain the expression for f2 and f3.
(b) State the values for which the functions f2 and f3 are not defined.

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