Precalculus Question for my test tomorrow

oh fuck, it didn't post the superscript.

It should be

"x^2 + y^2= 4

y^2 = 4 - x^2

√y^2 = √4 - x^2

y = ± √4 - x^2"

..

"y =√16 - x^2

So in this first case, it's pretty easy. Y is not a function of x because there's two possible answers for 4 - x^2 . -2^2 or 2^2 both equal four. In the second example, the textbook says that:

y =√16 - x^2"

>>44265
Can you just take a picture of the page in the textbook?

>>44270
Yeah one sec.

File: 2b_24.png (1KB, 200x112px) Image search: [Google]

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This is the equation.

Is y a function of x?

The answer in the back of the book says it is, but that doesn't make any sense to be since the other example is considered not a function.

File: wtf.gif (3KB, 352x353px) Image search: [Google]

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This is the other example. It's considered NOT a function.

>>44277
The other one, with the +/-, is not a function because there are two possible values of Y for given value X. It doesn't matter if two values of X would give the same Y, there just can't be two Y's for one X.

>>44280

That I know.

What I don't understand is that y =√16 - x^2 is considered a function.

It reduces to y = +/- √16 - x^2 , doesn't it?

File: plzhalp.jpg (4KB, 221x116px) Image search: [Google]

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Can no one halp me :(

>>44285
No. If you were to graph >>44279, it would look like a full circle of >>44277. This means that for a value of x, y can be two values.

>>44294

Can you explain more? I'm lost.

Sorry.

>>44297
A function is when there is only one value of y for a given value of x like>>44280 said. If you were to break y = +/- √16 - x^2 down, it would be y=+√16 - x^2 and y=-√16 - x^2. If you graph these two equations, it looks like a full circle with the positive equation being the top half like >>44277. To test the x has only one value of y, graph the equations and draw vertical lines. If you hit more than one point for each vertical line you draw, it is not a function. Imagine drawing a vertical line through a circle. You will always hit two points therefore it is not a function.

File: stilldontgetit.jpg (4KB, 115x115px) Image search: [Google]

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>>44303

But my textbook says y = √16 - x^2 is a function. :(

>>44305
Yes, and there's only one resulting y for each x, producing the semi-circle in the graph. The function doesn't reduce to +/- because you're not square rooting the y.

>>44305
Oops. Just replace all my 16 with 4.

>>44308

I'm confused.

I understand that x (independent variable) cannot have multiple outputs for y (dependent variable) simply because, in order for y to be a function of x, a domain (y) has to have a one to one ratio of Y to X. Having more than one value would mean that an element in the domain (y) is ascribed to more than one element in x, and therefore isn't a function.

What I'm not getting is why these two are basically the exact same problem but one is a function and one isn't. What does not square rooting the y have to do with the fact that there's multiple values for x? I was always taught if there were multiple possible values for x then it isn't a function of y.

>>44313
Are you trying to say that x is not a function of y or y is not a function of x?

>>44317

The question is to determine whether the equation represents y as a function of x.

Search Results

A function obviously defines one variable in terms of another. The statement "y is a function of x" (denoted y = y(x)) means that y varies according to whatever value x takes on, so y is the dependent variable.

That's why point. If x has two values it can't be the case that y if a function of x because then it violates the definition of a function.

>>44313
You're getting it backwards.
Multiple values of X can result in the same Y.
What you CAN'T have is a function where one value of X can give multiple values of Y.
This is the basis of the vertical line test. You're choosing one X, and if that X touches more than one value of Y on the graph, it's not a function.

File: Untitled.png (17KB, 673x221px) Image search: [Google]

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>>44321
Additionally, no, y =√16 - x^2 does not reduce to the +/-. Pic related.

>>44324

Ok, let me make sure I'm understanding this correctly.

So what you're saying is that, in the case where y is a function of x, you can have different values for x so long as they result in the same y.

And because in the first example (x^2+y^2 = 4) we had to square root the y AND the x, we get plus or minus because we also squared y. In the case of y = √16 - x^2, we can have several x values but because we don't square root the y it's okay?

Am I understanding you clearly?

I appreciate the graphical test that you're mentioning, but we haven't gotten to that part in the class yet I want to be able to just know from the equation without drawing a graph. So far in class we've only gone over doing it with equations and that's what we'll be tested on.

>Additionally, no, y =√16 - x^2 does not reduce to the +/-. Pic related

Sorry if I'm still really confused but the picture shows that the answer is +/-.

>>44331
>Sorry if I'm still really confused but the picture shows that the answer is +/-.
Which picture?

>>44332

The picture this dude posted:

>Additionally, no, y =√16 - x^2 does not reduce to the +/-. Pic related.

In his example, it shows an equation that's reduced to a +/- answer.

I still don't see how my first example and second problem are distinguishable besides the fact that the textbook tells me the second is somehow a function. I still fail to see why. The book says this can be calculated algebraically.

>>44331
To the first part, yes: multiple values of X can result in the same Y.

The difference between the circle and the semi-circle function given is the whole solving for y bit.
y =√16 - x^2 is already solved for y; there's no reduction. Putting in +/- before the root symbol is adding in other information. As depicted, with y solved as is, you're only dealing with the positive roots (The domain of sqrt(x) is |x|). Not having to square root the y DOES make that difference.

>>44342

I think this about clears it up.

So when we look at...

x^2 + y^2= 4

y^2 = 4 - x^2

√y^2 = √4 - x^2

y = ± √4 - x^2

What you're telling me is that we have a ± there because we had to calculate the square root of y, and that could have originally been a positive or negative number. Since we don't know, our answer is ±. This means that there are multiple y values for x, and is therefore not a function.

In the case of :

y =√16 - x^2

y= 4 - x

y is a function of x and there's no ± because we have y as is (we didn't have to square root it).

Which is a function.

Am I understanding you clearly?

Right when I thought I was getting it anon disappears.

Fuck I don't wanna fail my test.

>>44346
y =√16 - x^2 does not equal y= 4 - x.

>>44354

Why not ?

>>44360
Your algebra is wrong. You can't just take the roots of each term unless it is purely multiplication or division. If what you said was true, that would mean (4-x)^2 is y =√16 - x^2 but it ends up being x^2-8x+16

>>44363

Yeah I'm pretty fucked. It's a shame I can't get a hold of my professor during the week to help me with this shit considering I'm paying over $1000 for one fucking class and have to rely on 4chan, youtube, google, and the textbook to learn how to do this shit .

It's okay, he has tenure so he has no accountability. He'll continue to have classes averaging a 30% and still make like $100,000 a year for his great contribution to society.


Anyway, so the answer becomes

y = (-x+4)(x+4)

NOT y = 4 - x


Is my understanding of why this example is a function and the first example is not a function correct?

>>44368
No. Stop reducing y =√16 - x^2. There is nothing to reduce. You're still not even doing it right. If you were to reduce the inside of the root, it becomes y=sqrt[(x+4)(-x+4)].

>>44346
You need to distinguish two things:

If the square root is written in a formula, then it is a function!

b = sqrt(a) with a >= 0 and values b >= 0 (that is, sqrt(a) is never negative).

So in the second example y =sqrt(16 - x^2) is a function. It is only defined if 16 - x^2 >= 0, that is, -4 <= x <= 4, but in this interval the y-value is uniquely determined.

In the first example you take the square root to determine ALL solutions of the equation
y^2 = 4 - x^2.

You end with two possible values for y for each x with -2 < x < 2 (namely y = ± sqrt(4 - x^2) ), exactly one solution for x = ±2 (namely y = 0), and no solution for |x| > 2.

This means that the equation does not represent y as a function of x (if you consider all possible values of x, which is implied).

>>44378


I accidentally left out the square root over (-x+4)(x+4). Thanks for this though.

>>44382

>If the square root is written in a formula, then it is a function!

Wait, really? Every time?

b = √ a

where a ≥ 0 and b ≥ 0

So in the second example y = √(16 - x^2) is a function. It is only defined if 16 - x^2 >= 0, that is, -4 <= x <= 4, but in this interval the y-value is uniquely determined.

This is where I'm getting confused again. Since we don't know the value of x, couldn't it just be a value that leaves this equation undefined or undetermined?

>>44387
>Wait, really? Every time?
Every time. That's why you write ± sqrt(...). It does not mean "the positive value of the square root" and "the negative value of the square root", it means "take the value of the square root (it's always >=0) and multiply it by ±1".

>Since we don't know the value of x, couldn't it just be a value that leaves this equation undefined or undetermined?

Yes. If the value of x is not determined then you usually take all possible values, but you may get a formula with no possible solution, for example y = 1/(x -x). In this case y is undefined for every x since you would always divide by 0. The domain of the function (the set of values in which the function is defined) is empty.

For y = sqrt(x) the domain is the set of all real numbers x >= 0 since you can't take the square root of negative real numbers.

In the case that the function is not defined for an x, you can either remove it or fix the value.

For example, y = x/x is equal to 1 for any x except 0, but y is not defined for x = 0. In this case it would be natural to set the value of the function to 1 also for x = 0:

y = x/x for x not equal to 0, y = 1 for x = 0
or simply
y = 1 for any real x.