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>> P.4. Odd and Even Functions
P.3. Composition of Functions
An important operation of functions is composition. Given two functions f and g, we may form new functions f◊g and g◊f defined by f◊g(x) = f(g(x)) and g◊f(x) = g(f(x)). In general, f◊g and g◊f are different, though there are important exceptions.
Example 1: Consider the functions f(x) = 2x and g(x) = x2 + 5. What are f◊g(x) and g◊f(x)?
Solution: We compute f◊g(x) = f(g(x)) = 2g(x) = 2(x2+5) = 2x2 +10 and g◊f(x) = g(f(x)) = f(x)2+5 = (2x)2+5 = 4x2+5. Note that these functions are not equal.
Example 2: Consider the functions f(x) = 2x2 - 1 and g(x) = 4x3 - 3x. What are f◊g(x) and g◊f(x)?
Solution: We compute f◊g(x) = f(g(x)) = 2g(x)2 - 1 = 2(4x3-3x)2 - 1 = 2(16x6 - 24x4 + 9x2) - 1 = 32x6 - 48x4 + 18x2 - 1 and g◊f(x) = g(f(x)) = 4f(x)3 - 3f(x) = 4(2x2-1)3 - 3(2x2-1) = 4(8x6 - 12x4 + 6x2 - 1) - 6x2 + 3 = 32x6 - 48x4 + 24x2 - 4 - 6x2 + 3 = 32x6 - 48x4 + 18x2 - 1. Note that these functions are equal.
Example 3: Consider the functions f(x) = 1 / (1 - x) and g(x) = 1 - 1/x. What are f◊g(x) and g◊f(x)?
Solution: We compute f◊g(x) = f(g(x)) = 1 / (1 - g(x)) = 1 / [1 - (1 - 1/x)] = 1 / (1 - 1 + 1/x) = 1 / (1/x) = x and g◊f(x) = g(f(x)) = 1 - 1 / f(x) = 1 - (1 - x) = 1 - 1 + x = x. Note that these functions are equal.
Functions can also be composed with themselves, as the following example shows:
Example 4: Consider the function f(x) = 1 / (1 - x) and g(x) = 1 - 1/x. What are f◊f(x) and f◊f◊f(x)?
Solution: We compute f◊f(x) = f(f(x)) = 1 / (1 - f(x)) = 1 / {1 - [1 / (1 - x)]} = 1 / [-x / (1 - x)] = (x - 1) / x = 1 - 1/x and f◊f◊f(x) = f(f◊f(x)) = 1 / (1 - f◊f(x)) = 1 / [1 - (1 - 1/x)] = 1 / (1/x) = x.
