Evaluating Composite Functions

Date: 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