Possible Values of f(0)

Date: 04/19/2002 at 12:33:31
From: Jody Klenetsky
Subject: Function

Question from a sample ACT exam:

If f(x+y) = f(x)f(y), what is (are) the possible value(s) of f(0)?

The choices given are 0, 1, 0 and 1, any positive real number, any 
real number.

The answer is 1, but I have not been able to get an explanation for 
that answer.

Can you help?

Date: 04/19/2002 at 13:05:49
From: Doctor Peterson
Subject: Re: Function

Hi, Jody.

When I don't know what to do, I start playing to see what happens. So 
what happens if y = 0?

    f(x+0) = f(x)f(0)

    f(x) = f(x)f(0)

Hmmm ... if this is true for all x, then f(0) has to be 1.

Unless ... if it happens that f(x) = 0 for all x, then f(0) doesn't 
have to be 1. So I think they were wrong! The answer should be "0 and 
1," though the former is a very special case.

A more complete way to say what I found is to continue my equations:

    0 = f(x)f(0) - f(x)

    0 = f(x)(f(0) - 1)

    f(x) = 0 for all x, or f(0) = 1

    If f(x) = 0 for all x, then it is true that f(x+y) = f(x)f(y),
    so this is a valid solution.

    If f(0) = 1, then regardless of what f(x) is, f(x+0) = f(x)f(0).
    This is in fact a possible solution because f(x) = a^x works, for
    any non-zero a, and f(0) = a^0 = 1.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/21/2002 at 22:43:15
From: Chad Pals
Subject: Functions

Recently I gave my students an ACT practice test with a problem type I 
have never seen before. It stated that for all x and y, f(x + y)
= f(x)*f(y). It then asked for what f(0) = .  The answer came out to 
be 1.  

I don't understand why it is one, or how one would solve this.

Date: 04/21/2002 at 23:02:05
From: Doctor Paul
Subject: Re: Functions

Here's how to do the problem:

We are told that f(x+y) = f(x)*f(y) for all x and y.  Thus it 
certainly must be true for the specific case when x = 0.

Then f(0 + y) = f(0)*f(y)

which forces f(0) = 1.

It's really quite an easy problem if you see how to do it. But the 
'if' is what they're testing. All of the problems are relatively easy 
if you see the trick. The tests are supposed to measure mathematical 
thinking ability. Whether or not they do is the subject of ongoing 
debate...

I hope this helps.  Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/