Composition
of Functions: Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition
( f
o
f )(x) = f ( f (x))
(g
o
g)(x) = g(g(x))
Sometimes you have to be careful with the domain and range of the composite function.
Since f (x) involves a square root, the inputs have to be nonnegative. This means that the domain (the set of xvalues) for f (x) is "all x > 0". Then, in (g o f )(x), where I'm plugging x first into f (x) = sqrt(x), the domain is at least restricted to "all x > 0". Let's see what the two compositions look like: ( f
o
g)(x) = f (g(x))
The domain for the square root is all inputs that make "x – 2" nonnegative. That is, all x such that x – 2 > 0. Solving this for x, I get that the domain of ( f o g)(x) is "all x > 2". Now to do the other composition: (g
o
f )(x) = g( f (x))
The domain for this is all inputs that make the square root defined. Since there is only "x" inside the square root, then: the domain
of (g o
f )(x) is "all
x >
0". If your initial functions are just plain old polynomials, then their domains are "all x", and so will be the domain of the composition. It's pretty much only if your dealing with denominators (where you can't divide by zero) or square roots (where you can't have a negative) that the domain ever becomes an issue. Usually composition is used to combine two functions. But sometimes you are asked to go backwards. That is, they will give you a function, and they'll ask you to come up with the two original functions that they composed. For example: Copyright © Elizabeth Stapel 20022011 All Rights Reserved
This is asking you to notice patterns and to figure out what is "inside" something else. In this case, this looks similar to the quadratic x^{2} + 2x – 3, except that, instead of squaring x, they're squaring x + 1. In other words, this is a quadratic into which they've plugged x + 1. So let's make g(x) = x + 1, and then plug this function into f (x) = x^{2} + 2x – 3: ( f
o
g)(x) = f (g(x))
Then h(x) may be stated as the composition of f (x) = x^{2} + 2x – 3 and g(x) = x + 1.
Since the square root is "on" (or "around") the "4x + 1", then the 4x + 1 is put inside the square root. I need to take x, do "4x + 1" to it, and then take the square root of the result: g(x) = 4x + 1, f(x) = sqrt(x), and h(x) = ( f o g)(x). << Previous Top  1  2  3  4  5  6  Return to Index Next >>



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