Section 6.1 Composite Functions

OBJECTIVE 1

Suppose that an oil tanker is leaking oil and we want to be able to determine the area of the circular oil patch around the ship. It is determined that the oil is leaking from the tanker in such a way that the radius of the circular oil patch around the ship is increasing at a rate of 3 feet per minute. A(t) = (3t)2 = 9t2

= f(g(1)) = f(4(1)3 + 1) = f(5) = 2(5)2 + 3 = 53 = g(f(1)) = g(2(1)2 + 3) = g(5) = 4(5)3 + 1 = 501 = f(f(-2)) = f(2(-2)2 + 3) = f(11) = 2(11)2 + 3 = 245 = g(g(-1)) = g(4(-1)3 + 1) = g(-3) = 4(-3)3 + 1 = -107

OBJECTIVE 2

f(g(x)) = f(4x + 1) g(f(x)) = g(x2 + 3) = (4x + 1)2 + 3 = (16x2 + 4x + 4x + 1) + 3 = 16x2 + 8x + 4 g(f(x)) = g(x2 + 3) = 4(x2 + 3) + 1 = 4x2 + 12 + 1 = 4x2 + 13 Domain for f, g, and f ° g is all reals Domain for f, g, and g ° f is all reals

Notice that the domain of g is { x| x ≠ 2 } so we exclude 2 from the domain. Domain of f is { x | x ≠ -4 } so solve for g(x) = -4 We must also exclude 1 from the domain. So the domain of f ° g is {x | x ≠ 2, x ≠ 1 } Note: x ≠ 1

The domain of g is {x | x ≥ 1} and the domain of f(g(x)) excludes x = 1. Therefore the domain of f °g is {x | x > 1 } The domain of f is {x | x ≠ 0 } and the domain of f(g(x)) will also be {x | x ≠ 0 }.