yeah.
I was trying with logarithms :rolleyes:
Oskilian
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yeah.
I was trying with logarithms :rolleyes:
Oskilian
2 and 4 are correct. Now whats the second set of numbers?
>>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 XY 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.
x = 4Quote:
Originally posted by Thantos
2 and 4 are correct. Now whats the second set of numbers?
y = 2
really, thats trivial
/\

That the same one as mine first answer
x=2
y=4

engineer223
No reread 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.
Is it 2 and 4
Yeppers
a circle is not a function, it's an equation, but not a function
Oskilian
y = sqrt(x)
It passed the vertical line test so it is a function
Next time, make "y" alone before calculating.

Engineer223
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?...) ...
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...
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
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 nonfunctionQuote:
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
YES!!! urrrggghh....
you see if you start with
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)
YOU DON'T START WITH y^2=x, ok?
Do I have to graph it?
Proof
y=sqrt(x)

Engineer223