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### Sometimes, Always, or Never True?

```
Date: 02/12/2002 at 19:02:16
From: Jon
Subject: Functions

I can find no explanation for this question.

Tell whether this statement is always, sometimes, or never true:

f(g(x)) = g(f(x))
```

```
Date: 02/12/2002 at 23:05:39
From: Doctor Ian
Subject: Re: Functions

Hi Jon,

Suppose f(x) = x + 1, and g(x) = x^2.  Then

f(g(x)) = g(x) + 1

= x^2 + 1

and

g(f(x)) = (f(x))^2

= (x + 1)^2

= x^2 + 2x + 1

So in this case, f(g(x)) is clearly not equal to g(f(x)), except when
x happens to be zero. So we know that it's not _always_ true.

On the other hand, suppose f(x) = x + 1 and g(x) = x - 1.  Then

f(g(x)) = g(x) + 1

= (x - 1) + 1

= x
and

g(f(x)) = f(x) - 1

= (x + 1) - 1

= x

So for these functions, f(g(x)) = g(f(x)), regardless of what the
value of x happens to be.

This is easiest to think about if you think of functions as little
processes. Then the composition of two processes says: Put something
into the inner function; and whatever comes out, use that as input to
the outer function.

The functions don't have to be mathematical. For example, if f(x)
means 'paint x blue', and g(x) means 'wrap x in paper', then

f(g(x)) => wrap x in paper, and paint the whole thing blue

but

g(f(x)) => paint x blue, and wrap the painted object in paper

In the first case, the paper gets painted; in the second, the object
itself gets painted.

In many cases where f(g(x)) = g(f(x)), the functions are inverses of each
other.  That is, what one does, the other undoes, so when you compose
them (in either order), you get x back as the final result.  You've
seen one example.  Here are some others:
__
f(x) = x^2, g(x) = \| x

f(x) = 10^x, g(x) = log(x)

f(x) = 2x, g(x) = x/2

f(x) = sin(x), g(x) = arcsin(x)

But this isn't always the case, e.g.,

f(x) = 1/x,  g(x) = -x

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Functions

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