Evaluating Composite FunctionsDate: 1/31/96 at 10:13:35 From: Anonymous Subject: Calculus What is the domain of f(x) = 3? Also, if f(x) = 3, and g(x) = x^2 + 1 divided by the square root of (x^2 -1), how do you evaluate the composite function of f composed with g (f circle g)? The constant really threw me off in these problems (am I making a simple thing hard?). Date: 6/18/96 at 0:48:20 From: Doctor Paul Subject: Re: Calculus The domain of a function is the set of points on the x-axis that can be plugged into the equation with a finite result returned. In other words, if a particular x leads to an undefined point, then that point is not in the domain of f(x). For example: x if f(x) = ----- (x-3) Then the domain would be the set of all x's (from -infinity to positive infinity) with the exception of x=3 (b/c x=3 leads to dividing by zero). Since your function, f(x) = 3 doesn't have any x's in it (and is hence just a horizontal line) the domain is the set of all x's (from -infinity to infinity) The composite function, f(g(x)) [pronounced "f of g of x"], is very simple in this case. Let me explain the concept first. If f(x) = x^3+1 and g(x) = x^2-4 Then f(g(x)) is calculated by plugging g(x) directly into every occurring x in f(x). In this case, f(g(x)) = (x^2-4)^3 +1. In your case, since the function f(x) is just a constant, there is nowhere to plug g(x) into f(x). Thus, f(g(x)) is just 3. In the case where f(x) is a constant, f(g(x)) is always a constant regardless of how complicated g(x) is. -Doctor Paul, The Math Forum |
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