Math Question, hard

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• 12-03-2001
oskilian
yeah.

I was trying with logarithms :rolleyes:

Oskilian
• 12-04-2001
Thantos
2 and 4 are correct. Now whats the second set of numbers?
• 12-04-2001
y^2 = x is a function
>>no it isn´t, a function must have only one X value for each Y >>value, and this one doesn´t (for example: if X=4, y can be 2 >>or -2)

Osk, you're trying to writ some code where you need to understand 3D. OK, that's fair. But, then admit that you can have more than one value of each X and Y.

For example, let's say I want to define a circle in 3D (even the circle has only 2D, it's in a space with 3D):

1) x^2 + y^2 = 1
2) z = 0

The ecuation 1) defines the limits (I don't know how to express this in Enlgish) of the function in the X-Y plane, and the function 2) express the position of the circle in the Z axis.

And the function Y^2 = x is used many times when you need a 3D boundary to calculate for example triple integrals, which give you the volume of a body.

By the way, I took this examples from my first year at uni (5 or 6 years ago I think).

And these functions are quite simple, because a value of x has only two values of y; there are functions that have more y values for each x value. Like the Rosae Function, this function draws like a rose, but I don't know if you call it like this in English as well. If I find something in my notes I'll post it.
• 12-04-2001
Vulcan
Quote:

Originally posted by Thantos
2 and 4 are correct. Now whats the second set of numbers?
x = 4
y = 2

really, thats trivial
• 12-04-2001
Yoshi
/\
|
That the same one as mine first answer

x=2
y=4
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engineer223
• 12-04-2001
Thantos
No re-read my comment. I said 2 and 4 were correct. But there is another correct set of numbers. Meaning that 2 and 4 isn't the only set of correct numbers.
• 12-04-2001
maes
Is it -2 and -4
• 12-04-2001
Thantos
Yeppers
• 12-04-2001
oskilian
Re: y^2 = x is a function
a circle is not a function, it's an equation, but not a function

Oskilian
• 12-04-2001
Yoshi
y = sqrt(x)
It passed the vertical line test so it is a function

Next time, make "y" alone before calculating.
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Engineer223
• 12-04-2001
*Michelle*
GEEZ! Engine, you don't get my point...

> y˛ = x is a function
sqrt both sides

y = sqrt(x)
It passed the vertical line test so it is a function<

That was exactly what I said, instead I said it the other way around... You just proved my point, y sqrt(x) is a function while y^2=x is exact same equation but not a function. ONLY y = sqrt(x) is a function the other one is not. The definition for a function, is that it's linear graph thing is not touched twice by a vertical line. And we are talking about 1D here ( is there such a thing as 1D?...) ...
• 12-05-2001
doubleanti
the thing with your examples is whether to decide to take the negative portion of a sqrt... and it's just a matter of orientation...
• 12-05-2001
oskilian
you must undertand that you cannot do that!

the inverse function of the second power IS NOT int sqare root!!!

graphing y^2=x will give you a different graph than y=sqrt(x)

now this is true:

y^2=x
y=(+/-)sqrt(x)

but neither of them are functions, let me copy the "function" definition from my Calculus book

Quote:

Originally written by Earl W. Swokowsky
An f function of a D group over an E group is a correspondence which assigns each element x of D one unique element y of E
one UNIQUE element; in the equation y^2=x, y=2 and y=-2 are assigned to x=4. There are two solutions for the eqation, and THAT makes y^2 a non-function
• 12-05-2001
*Michelle*
YES!!! urrrggghh....

y=ROOT(x)

that IS a function.

but if you move the root sign over, you will get y^2=X

which IS NOT a function. (because x will be both -2 and 2)