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Homework Help: Which of the following represent a function

  1. Jun 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Which of the following represent a function:
    a)[tex]y=-1/2x+3[/tex]

    b)[tex]x^2+y^2=25[/tex]

    c)[tex]y=2x^2+7x+3[/tex]

    2. Relevant equations


    3. The attempt at a solution
    a)
    b) is a circle so has 2 values for y and is not a function.
    c)
    I know that functions only have 1 y relation to x but don't know how to prove whether a and c are functions or not.
    Thank you
     
    Last edited: Jun 4, 2015
  2. jcsd
  3. Jun 4, 2015 #2
    If domain and codomain of the functions are f: R-->R . Then "a" can't be function because there is no answer for x=0. And also like you said b can't be a function because there are two reflections of x=a.
     
  4. Jun 4, 2015 #3
    Hi Thermo,
    Thank you for your reply,
    Others? meaning b and c?
    I don't agree or I am missing the plot.
    B cannot be a function because it is a circle...
     
  5. Jun 4, 2015 #4
    Yes I realised it later sorry for that. So the answer must be only c is a function. However domains and codomains matter in this case.
     
  6. Jun 4, 2015 #5
    Would I be able to solve the equation to prove this?
     
  7. Jun 4, 2015 #6

    SammyS

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    By the way: Thermo is incorrect.

    What you have for (a) literally means ##\displaystyle\ y=-\frac{1}{2}x+3\ .\ ## Is that what you mean?

    Or do you mean ##\displaystyle\ y=-\frac{1}{2x}+3\ ?\ ##

    You can graph the equations , (a) and (c) .

    You should be able to identify what figures the graphs produce.
     
  8. Jun 4, 2015 #7
    Hi Sammy,
    you are correct,
    I get a u shape for c
    and upside down u shape for a.

    So a and c are functions.
    Thank you so much for your input.
     
  9. Jun 4, 2015 #8

    SammyS

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    That U shape is called a parabola.

    What is the graph of equation (a) called ?
     
  10. Jun 4, 2015 #9
    Sorry for the wrong info I wasn't so sure I think I am mistaken for continuity or differentiability.
     
  11. Jun 4, 2015 #10
    a Hyperbola
     
  12. Jun 4, 2015 #11
    no problem, Luckily there are many smart guys and gals to double check one another.
     
  13. Jun 4, 2015 #12
    there is a second part to the question:
    which statements do not define a one-to-one function?

    My answer is B because it is a circle and not a function.
     
  14. Jun 4, 2015 #13

    cnh1995

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    Basic equation of a function is y=f(x).. Aren't all of them functions?? B isn't a one to one function and A isn't a continuous function.. They are just the types of function..
     
    Last edited: Jun 4, 2015
  15. Jun 4, 2015 #14
    From my text book:
    A function f between two sets of real numbers A and B is a relation in which each element of A is paired with a unique element of B.

    If you draw a vertical line on the graph, is it possible that the line can intersect the graph at 2 places? If it does the equation is not a function.
     
  16. Jun 4, 2015 #15
    Although,
    The function y = f(x) is a function if it passes the vertical line test. It is a one-to-one function if it passes both the vertical line test and the horizontal line test.

    Then none of these are a 1-1 function.

    Could someone agree/disagree with this?
    Thank you,

    Jaco
     
  17. Jun 4, 2015 #16
    A and C because B is not a function.
    I am not sure anymore.
     
    Last edited: Jun 4, 2015
  18. Jun 4, 2015 #17

    cnh1995

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    Well is it due this "precalculus" concept? Because as per my knowledge, these are all treated as functions in calculus.
     
  19. Jun 4, 2015 #18

    SammyS

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    These three equations might be treated in Calculus, but equation (b) better not be referred to as a function.
     
  20. Jun 4, 2015 #19

    Mark44

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    The first one, which I am assuming is y = -(1/2)x + 3, is a function, and is one-to-one.
     
  21. Jun 4, 2015 #20

    Mark44

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    That is just function notation, but isn't any sort of basic equation.
    No, not all of them are functions.
    The equation in a) is a continuous function. In clearer form, the equation is y = -(1/2)x + 3.
     
  22. Jun 4, 2015 #21

    Mark44

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    Sammy is correct.
     
  23. Jun 4, 2015 #22
    I suppose it is 1/(2x)
     
  24. Jun 4, 2015 #23

    Mark44

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    I believe it was meant as (-1/2)x, not (-1)/(2x). @Jaco hinted that my thought is correct, but he didn't explicitly say that.

    In any case, what he wrote was this: y=−1/2x+3. The correct interpretation of this is (-1/2)x + 3, not -1/(2x) + 3.
     
  25. Jun 5, 2015 #24
    (-1/2)x + 3, not -1/(2x) + 3
    as mark says,

    Sorry for the typo, still getting used to typing like this.
     
  26. Jun 5, 2015 #25

    SammyS

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    In that case, when you said the graph for equation (a) was a hyperbola, you were incorrect.
    which was in response to
    Equation (a) : ##\ y=(-1/2)x + 3\ ## gives the graph of a line with slope of ##\ -1/2\ ## .

    By the way: Did you say that none of the equations gave a 1 to 1 function ?
     
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