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Date: 04/02/2002 at 14:08:23
From: Teresa Otworth
Subject: Range

If f(x) = x-3/x+3, I know the domain to be all Reals except for -3.  
The range, however, is all Reals except for 1. I don't know how they 
get the 1 algebraically.  

Thank you for your time. 
Teresa Otworth

Date: 04/02/2002 at 17:08:26
From: Doctor Peterson
Subject: Re: Range

Hi, Teresa.

The usual way to do this is to try to invert the function; then the 
domain of the inverse is the range of the function. That is, we solve 
for x:

    y = (x-3)/(x+3)

    y(x+3) = x-3

    xy + 3y = x - 3

    xy - x = -3 - 3y

    x(y - 1) = -(3 + 3y)

    x = (3 + 3y)/(1 - y)

Now we can see immediately that y can be anything but 1. Since for any 
y except 1, there is an x, we know that the original function can take 
any value except 1.

Having done that, we can look back at the original function and see 
why it can't be 1. If we had

    (x-3)/(x+3) = 1

then x-3 = x+3, which is impossible.

- Doctor Peterson, The Math Forum   

Date: 04/04/2002 at 13:59:29
From: Teresa Otworth
Subject: Range

In f(x) = 2/sqrt(x-4), how do you find the range?

Date: 04/04/2002 at 15:53:36
From: Doctor Peterson
Subject: Re: Range

Hi, Teresa.

Did you try using the method I showed you last time? We can help a lot 
better if you show us your work so we can see why you are having 
trouble - just as a medical doctor wants to see where it hurts, a math 
doctor wants to see what doesn't work.


    y = 2/sqrt(x-4)

for x, and see what values of y do not allow you to find an x. Those 
values are excluded from the range.

In this case there is an extra issue: the square root is always taken 
as the POSITIVE root, which limits y further.

- Doctor Peterson, The Math Forum   

Date: 04/04/2002 at 19:28:36
From: Teresa Otworth
Subject: Range

Thanks for being patient with me.  I worked f(x)= 2/sqrt(x-4) like 

   y(sqrt(x-4)= 2.  

Then, I moved the y:  

   sqrt(x-4)= 2/y.  

Then I squared both sides and obtained 

   x-4 = 4/y^2  

This is where my question arises.  If I then move the four on the left 
side of the equation and then solve for one y, I get an imaginary 
number.  Was that what you were referring to that would always be 
positive?  That would give a 4 + 4, so I am lost at that point.

Date: 04/04/2002 at 22:18:50
From: Doctor Peterson
Subject: Re: Range

Hi, Teresa.

You've done well so far. Remember you are solving for x, and you get

    x = 4 + 4/y^2

You don't want to solve for y now; the purpose of solving for x was to 
see for which values of y you can find an x. Look at this equation, 
and you see that you can put any value of y except for 0 into it; the 
x you get out is just the x you could put into the original equation 
to get that value of y. So f(x) can take any value but 0.

But ...

Let's try an example or two. If y = 1, we get x = 4 + 4/1 = 8, and

    f(8) = 2/sqrt(8-4) = 2/2 = 1

so we did in fact show that 1 is in the range. Now let y = -1. We find 

    x = 4 + 4/(-1)^2 = 4 + 4/1 = 8

But that's the same x we got for y = 1, and we know that f(8) is not 
-1. So what happened?

When you square y, both 1 and -1 give the same result. Another way to 
say that is that 1 and -1 are both square roots of 1. But in the 
original equation, we had the square root _function_, which is defined 
as the POSITIVE square root only. So

    f(x) = 2/sqrt(x-4)

always gives the positive root, and so f(x) has to be always positive. 
That's why we don't get -1 out of it when we put 8 in, even though 
both 2 and -2 are square roots of 4.

So the answer is that the range of this function is all positive real 
numbers: zero is excluded because we couldn't divide by y^2 in our 
inverse, and negative numbers are excluded by the square root 

I've gone through the whole thing, because this last part is tricky. 
Let me know if you don't follow any part of it.

By the way, a good way to get a better feel for these problems is to 
try graphing them after you've worked out the domain and range by 
other means. Do that in this case and you will see what's going on 
much more clearly.

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Functions

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