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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?

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

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/
```
Associated Topics:
High School Functions

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