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Range

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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.

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
http://mathforum.org/dr.math/
```

```
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.

Solve

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
http://mathforum.org/dr.math/
```

```
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
this:

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
that

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
function.

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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Functions

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