Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**handsonmath****Member**- Registered: 2011-03-04
- Posts: 6

y is a function of x

x = -y^2

A textbooks tells me the domain of x = -y^2 = (-infinity, 0] and the range is (-infinity, infinity)

But, its not clicking how they get this

I'm getting D[0,infinity) and R(-infinity, 0]

Read until you have something to write...Write until you have nothing to write...when you have nothing to write, read...read until you have something to write (http://handsonmath.blogspot.com/)

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi handsonmath;

Solve that equation for y and then do the domain and range, tell me what you get.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**handsonmath****Member**- Registered: 2011-03-04
- Posts: 6

-sqr(x) = y

I'm getting D[0,infinity) and R(-infinity, 0]

Read until you have something to write...Write until you have nothing to write...when you have nothing to write, read...read until you have something to write (http://handsonmath.blogspot.com/)

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I am getting to solve it.

But that is not important.

Domain = ( -∞, 0] and the Range is (-∞,∞)

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**TMorgan****Member**- Registered: 2011-04-13
- Posts: 25

Hello,

This is my first post here and I do not yet know how to make equations look good. I did however want to comment on this problem.

I believe there is a problem with the original premises. The equation

x = -y^2

and the statement

y is a function of x

are not, in my opinion, compatible. If you accept the definition of a function as a rule that assigns every element of one set (the domain) exactly one value of another set (the range), then in this equation x is a function of y. For every value of y (-1, 0 +1) there is exactly one value of x (-1, 0 -1). So x is a function of y. But for a value of x (-1) there are 2 values of y (-1, +1) so y is not a function of x.

As far as discussing what the correct values are for the domain and range it first must be agreed which variable (x or y) is a function of the other, which will determine which is the domain and which is the range. Also it must be clarified if the original equation

x = -y^2

is to be interpreted as

x = -(y^2) or

x = (-y)^2.

I think the way it is written implies the first, but the way bobbym solved for y, without putting a plus/minus before the square root sign, implies the second. I think either way y is still not a function of x.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi TMorgan;

Welcome to the forum! I solved it like this obviously.

The square takes precedence so - y^2 = - (y^2).

I assumed x >=0 so I did not put the plus minus in front. Anyway it was not necessary to solve it and did not want to confuse him with ± i.

You might be right about the lack of rigor of that problem but I rarely argue with their teachers or their textbooks. I just try to guess what they want.

This is my first post here and I do not yet know how to make equations look good.

Here is where you go to start learning latex.

http://www.mathisfunforum.com/viewtopic.php?id=4397

If you are lazy like me then you can go here and this page will do it for you.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,462

hi handsonmath

Before coming to the domain range difficulty just have another look at

-sqr(x) = y

It should be sqr(-x) = y

Now you can choose a domain for a function. It is not pre-defined by the function.

For example

y = 2x. I can have domain = {counting numbers} or domain = {integers} or domain = {real numbers} etc etc.

So why would a book say

domain of x = -y^2 = (-infinity, 0]

Well if x is positive then the square root takes you into complex numbers. Nothing wrong with that but your book has chosen not to. By picking any negative number and zero, the book makes sure that y is real.

Now about the range.

If I define a function as y = sqr(-x), then any mathematician is entitled to ask, which square root am I supposed to take (+ or -) because a function must be well defined ... that is to say, have just one value of y for any x we choose.

But the function you started with isn't defined like that.

It is defined 'backwards' ie. x = -y^2

Does that make any difference?

So we choose an x in the domain (let's say x = -9) and try to find a y to fit. I can find two values that work y = + 3 and y = -3. So I guess that's why the book wants the range to be -infinity to + infinity.

But I'm with you on this.

The book range means it isn't a well defined function as it has two values of y for any x (not equal to zero).

I like your range better

R(-infinity, 0]

But you are implying that the negative square root should be taken. Again, nothing wrong with that, but some people might wonder why you didn't have (0,+infinity]. This is also possible until we get clarity on the square root.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

Pages: **1**