Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: HELP FAST y^(2)=x+1 IS THIS A FUNCTION?

  1. Dec 28, 2004 #1
  2. jcsd
  3. Dec 28, 2004 #2
    (square root of x) +1 is also a function right!
  4. Dec 29, 2004 #3
    [tex] y^2 = x +1 [/tex]
    [tex] y= \sqrt(x+1) [/tex] <- is a function of y in terms of the variable x.

    [tex] \sqrt(x) +1 [/tex] is not a function, there is no equal sign. Also, there has to be a dependant and independent term. There is an independent term x, but no dependent term y anywhere.
    Last edited: Dec 29, 2004
  5. Dec 29, 2004 #4
    Unfortunately mathematical formalism isn't often taught very well. There is also a lot of sloppy shorthand that goes around.

    To begin with no equation is truly meaningful unless the domain and range of the variables has been stated, along with any other conventions that might be in use.

    As given, the equation you wrote is technically not a function. However if you write it like so:

    [tex]y^2=x+1 ; x\geq -1; y\geq 0[/tex]

    Then it qualifies as a function. :biggrin:

    However most people would write it as [itex]y=\sqrt{x+1}; x\geq -1[/itex] and then just say that they are only considering the positive roots.

    Without these limitations it would not be a function, at least not if [itex]x[/itex] and [itex]y[/itex] are taken to be real numbers.
  6. Dec 29, 2004 #5
    how come my teacher says that y^2=x+1 is not a function is that fair?

    Isnt that a trick question how are u supposed to know whether to rearrange it or not? Normally it would be written y= not y^2=
  7. Dec 29, 2004 #6

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's fair if it was explained to you at the outset that when the domain and/or range are not given, then you are supposed to assume the implied domain of definition of the relation. In that case, you can either graph the relation to see if it passes the vertical line test (it doesn't) or plug in a couple values of x in the domain and solve for y. If you get more than one value of y for even a single value of x, then the relation is not a function.
  8. Dec 29, 2004 #7
    y^2=x+1 is not a function.
    For each value of x, there are two value of y. ( except x=-1)
    Hence, this is not the definition of function.
  9. Dec 29, 2004 #8

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member


    I know that I've mentioned this in one of your threads before, but I'm saying it again now with emphasis: You have to show how you started the problem before receiving help here. Even if your start is wrong, you have to show what you did or what you think about the problem. Please remember that in the future.

    Thank you,
  10. Dec 29, 2004 #9
    Whether it's fair or not actually depends on what your teacher told you in lectures, and/or what's written in your textbook.

    Typically in math classes if it's written like:


    It's taken to be a function because the assumption or convention is that we only take the positive value of the square root. However, sometimes that convention isn't made clear. Also, this isn't always the convention if you came up with this relationship from a word problem! So it can be a source of real confusion for someone starting out in mathematics.

    On the other hand, if the relationship is given as:


    Then in order to rearrange this in terms of [itex]y=\sqrt{x}[/itex] we need to take the square root of both sides. Whenever we take the square root of both sides we need to keep in mind that there are both positive and negative roots. So we need to write the result of taking the square root of both sides as:

    [tex]y=\pm \sqrt{x}[/tex]

    This is different from [itex]y=\sqrt{x}[/itex].

    Is this fair? Well, sure it's fair. In one case you were given a conventional square root function which is taken by convention to be positive . In the other case you manipulated the relationship after it was given to you, and when you did that you need to consider the consequences.

    However, I honestly believe that math instructors could explain this a little better sometimes rather than springing it on everyone like a bad joke. :frown:

    I honestly believe math instructors don't often explain this stuff well in the early going. They kind of let the students discover these things along the way which isn't necessarily the best pedagogy IMHO. Explaining conventions early on is always a good idea I think.

    So this whole topic isn't so much about the rules of mathematics as it is about the style of pedagogy. I mean, mathematics is hard enough for most people to learn. Confusing the issue by assuming conventions or not making them clear in the early going doesn't help matters much. I have a really big bone to pick with a lot of math professors on the issue of pedagogic methodologies to be quite honest. :grumpy:
  11. Dec 29, 2004 #10

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It depends on the definition of function that you are using. Also, are you asking whether it is a 'function' of x?
    [itex]y^2=x+1[/itex] is certainly a function of y (f(y)=y^2-1). [itex]y^2=4px+c[/itex] is the form of function used to define a parabola with focus at p.

    Most of us would say that [itex]f(x)=\sqrt{x+1}[/itex] defines f(x) as a function of x, although [itex]\sqrt{x}[/itex] has two values for each x (+ and -).

  12. Dec 29, 2004 #11
    this can also be verified (but may be tedious) by graphing it... and you can say whether it is a function or not by vertical line test.
  13. Dec 29, 2004 #12
    You guys use way too complicated explanations. Think in context. This is the K-12 board. Anyway, it's a function when for one value of x you get one value of y. If you get multiple answers of y for one value of x than y is not a function. This can most easily be done by the verticle line test. If you graph the equation and the verticle line intersects with more that 1 point than it's not a function(at least in terms of y it's not a function.)

    As for you problem solve for y and see if x gives you 2 answers. Remember square roots have both a positive and negative answer.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook