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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?  


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 

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 

If you need more help, write back with a sample problem and your own 
work on it.

- Doctor Peterson, The Math Forum   

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
  x - 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
Associated Topics:
High School Functions

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