Domain of a Composite Function
Date: 08/30/2001 at 15:08:09 From: Michele Subject: Composite Functions I don't completely understand how to find the domain of a composite function. Could you please explain it? I realize that I have to consider the domain and range of the "inside" function and then see how they relate to the domain of the "outside" function, but I don't know how to find the correct domain every time. Can you help? Thanks.
Date: 08/30/2001 at 17:02:30 From: Doctor Peterson Subject: Re: Composite Functions Hi, Michele. There's no really simple trick to get it right every time; it might help to see an example you got wrong, so I could see what kind of mistakes you make and help you avoid those. Let's take an example: f(x) = sqrt(x-1) g(x) = 1/(x-1) fg(x) = sqrt(1/(x-1) - 1) The domain of f is the set of x for which f(x) is defined; since the square root is not defined for negative numbers, f is not defined for x<1. The domain is x>=1. The domain of g is the set of x for which x-1 is not zero; that is, all x except 1. Now look at fg. We can first exclude x=1 from the domain, since g itself is not defined there. But then we have to determine when 1/(x-1)-1 is negative: 1/(x-1) - 1 < 0 1/(x-1) < 1 x-1>0 AND 1 < x-1; OR, x-1<0 AND 1 > x-1 x>1 AND x>2; OR, x<1 AND x < 2 x>2 OR x<1 (The "and/or" stuff came in when I had to multiply by x-1; I can't do that unless I know whether it is positive or negative, so I need two cases.) So fg will be defined only when 1<x<=2, since x=1 has been excluded. This is its domain. Notice that I have not used the composite nature of fg explicitly in doing this; I've just worked with the function fg itself, like any other function. What did this have to do with f and g? First, the domain of fg must be a subset of the domain of g; that's why I had to exclude x=1. Second, the domain must include only those values of x which g maps into the domain of f; that's why I had to keep the argument of the square root non-negative. This is not something you can do just by looking at the domain and range of f and g; you have to solve g to find x for which g(x) is in the domain of f. Now what is the range of fg? The easiest way is often to invert it and find the domain of (fg)^-1. You will be doing the same sorts of things. If you need more help, write back with a sample problem and your own work on it. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 02/16/2006 at 02:39:58 From: Eddie Subject: Domain of a Composite Function Could I get an explanation of one of the steps in that answer? I don't understand the part with the AND and OR and the two cases once you get to 1/(x-1) < 1.
Date: 02/16/2006 at 09:56:29 From: Doctor Peterson Subject: Re: Domain of a Composite Function Hi, Eddie. What I was saying there was something like this: We can't multiply an inequality by a variable quantity without knowing whether the multiplier is positive or negative, so we have to break it into two cases: 1. If x-1 > 0 (ie, its positive), then we can multiply without changing the <: 1/(x-1) < 1 (x-1) * 1/(x-1) < (x-1) * 1 1 < x-1 2 < x x > 2 So one way in which the inequality can be true is if x-1 > 0 and x > 2 x > 1 and x > 2 which will be true whenever x > 2. 2. If x-1 < 0 (ie, its negative), then we have to change the < to >: 1/(x-1) < 1 (x-1) * 1/(x-1) > (x-1) * 1 1 > x-1 2 > x So another way in which the inequality can be true is if x-1 < 0 and x < 2 x < 1 and x < 2 which will be true whenever x < 1. Putting these together, the inequality will be true when x < 1 OR x > 2. I'm not sure why I did it that way; there's a neater way to solve this sort of inequality without having to resort to cases. We rewrite it as a rational expression being compared to zero, and then factor it: 1 ----- < 1 x - 1 1 ----- - 1 < 0 x - 1 1 x - 1 ----- - ----- < 0 x - 1 x - 1 2 - x ----- < 0 x - 1 The left side will be negative when EITHER 2-x is positive AND x-1 is negative, OR 2-x is negative AND x-1 is positive. A nice way to work this out is to make a chart using a number line, showing the sign of each factor in each region (less than 1, equal to 1, between 1 and 2, equal to 2, greater than 2): 1 2 <-------+---------+--------> 2-x: + + + 0 - x-1: - 0 + + + ratio: - u + 0 - (u = undefined, due to 0 denom) When is the ratio negative? When x < 1 or when x > 2, as we saw the other way. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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