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Math Help - How to know if a relation is a function

  1. #1
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    How to know if a relation is a function

    Hi, I'm having a serious headache here. I know about the vertical line test and x cannot be repeated. But how to tell if the equations below are a function of not without drawing a graph?

    1. sin(x+y)=1

    2. x + e^y = 1

    3. x = (sin y)^2

    4. y - x - x^(1/2) = 17

    5. y = (4x - 2)/(3x - 1)

    Any help is greatly appreciated. Thanks in advance.
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  2. #2
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    Apart from the vertical line test the only other way I know is committing the shapes of each function to memory

    x + e^y = 1

     e^y = 1-x

     y = \ln(1-x)

    This one looks like a funciton.
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  3. #3
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    Quote Originally Posted by chlorophyll69 View Post
    Hi, I'm having a serious headache here. I know about the vertical line test and x cannot be repeated. But how to tell if the equations below are a function of not without drawing a graph?

    1. sin(x+y)=1

    2. x + e^y = 1

    3. x = (sin y)^2

    4. y - x - x^(1/2) = 17

    5. y = (4x - 2)/(3x - 1)

    Any help is greatly appreciated. Thanks in advance.
    You can do a limited form of the vertical line test easily without graphing. For example, take sin(x+y)=1 and let x = 0. So sin(y) = 1. Now, we know there are multiple y values that satisfy this equation because sine is periodic. Therefore, this is not a function. (In case you didn't quite follow, we have the point (0,pi/2) on our graph, also (0,5pi/2), etc.)

    And you can know that all polynomials y=P(x) are functions, and in general solving for y can tell you things that might not otherwise appear obvious.
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  4. #4
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    A function can always be written in the form y= f(x) where f some unambiguous formula in x. By "unambiguous" I mean that one value of x can only give one value of y (which is really just saying "f is a function of x").

    For example, if I try to solve sin(x+y)= 1 for y, I will have to use the "arcsin" which is NOT a function without being restricted.
    x+ y= arcsin(1) so y= arcsin(1)- 1. That is not a function because arcsin is not a function- one value of x can give different vaues of arcsin(x): sin(0)= 0 so arcsin(0)= 0, sin(pi)= 0 so arcsin(0)= pi, etc.

    If x+ e^y= 1 then e^y= 1- x and y= ln(1- x). ln is a functions so x+ e^y= 1 defines a function.

    If x= sin^2 y then sin y= sqrt(x) and so y= arcsin(sqrt(x)). That's not a function because arcsine is not single valued.

    If y- x- x^(1/2)= 17, then y= x+ x^(1/2)+ 17. That's easy, that's a function.

    If y= (4x - 2)/(3x - 1) it's already written as "y= " and is obviously a function.

    Note an important difference: if y= arcsin(x), then it is assumed that "arcsin(x)" is referring to the "restricted" inverse of sin: arcsin(x) is between 0 and pi so single valued and a function. But if x= sin(y), then we cannot solve for y without getting multiple values. If y= sqrt(x) or y= x^(1/2), those are defined as functions- they are single valued: sqrt(4)= (4)^(1/2)= 2, not -2. But if y^2= x, then y= sqrt(x) or y= -sqrt(x), not a function.
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