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    State whether the following mappings are functions, and is they are one-one, one-many,many-one or many-many relations.

    a) x^2+y^2=1
    b) y=x^4
    c) y = √x (x ≥0)

    For the first mapping it is not a function but I am not sure what type of relation it is. In addition, for the second I think it is a one-many relation and the third I am not sure altogether.


    Then : find fg(x) and gf(x)
    When f(x) = x^2 and g(x) = 2x+3

    Would fg(x) be (2x+3)^2;
    because f(g(x)) then substitute the value of g into f

    Likewise gf(x) = 2x^2 +3
    because g(f(x)) then substitute the value of f into g= 2x^2 +3

    Finally,
    If f(x) =x^2+4 find the inverse function of f^-1(x).
    For this would I find the inverse of f^-1(x) or the first function?

    In either case the inverse function of f(x)=x^2+4 = ±√x-4
    Or for the second function f^-1(x) = fx

    Are these correct?
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    (Original post by AliceAM342)
    State whether the following mappings are functions, and is they are one-one, one-many,many-one or many-many relations.

    a) x^2+y^2=1
    b) y=x^4
    c) y = √x (x ≥0)

    For the first mapping it is not a function but I am not sure what type of relation it is.
    Correct, it's not a function. But note that a single y value (between -1 and 1) corresponds to two distinct x values (between -1 and 1). Similarly, a single x value corresponds to two y values. So the relation is...?

    In addition, for the second I think it is a one-many relation
    Not quite. A single x value corresponds to a single y value. But, a single y value corresponds to two x values.

    and the third I am not sure altogether.
    Same procedure, how many values does a chosen x output? How many values does a chosen y output?

    When f(x) = x^2 and g(x) = 2x+3 ... Would fg(x) be (2x+3)^2 ... Likewise gf(x) = 2x^2 +3
    These are correct.

    Finally,
    If f(x) =x^2+4 find the inverse function of f^-1(x).
    For this would I find the inverse of f^-1(x) or the first function?
    In either case the inverse function of f(x)=x^2+4 = ±√x-4
    That's fine.
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    (Original post by RDKGames)
    Correct, it's not a function. But note that a single y value (between -1 and 1) corresponds to two distinct x values (between -1 and 1). Similarly, a single x value corresponds to two y values. So the relation is...?



    Not quite. A single x value corresponds to a single y value. But, a single y value corresponds to two x values.

    RDKGames


    Same procedure, how many values does a chosen x output? How many values does a chosen y output?


    These are correct.



    That's fine.
    For the first mapping would it be a many-many relation because , as you said, there is a single y value to two x values and a single x value to two y values, this mapping has single y and x values so would the other value then correspond at two of the value?
    For the second mapping would it be a one-many relation , which is not a function,?
    And for the third mapping would it be a one-many relation again (not a function) because there is a singular y value and multiple x values?

    Thank you for your comments on the other questions.
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    (Original post by AliceAM342)
    For the first mapping would it be a many-many relation because , as you said, there is a single y value to two x values and a single x value to two y values, this mapping has single y and x values so would the other value then correspond at two of the value?
    Correct, but I don't understand your question.

    For the second mapping would it be a one-many relation , which is not a function,?
    Not quite.

    And for the third mapping would it be a one-many relation again (not a function) because there is a singular y value and multiple x values?
    Not quite. One-many means a single x outputs multiple y.
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    Would the second mapping be a many-one relation, because there is a single y value to 4 x values.
    For the third mapping would it be a many- one relation, because there is a single y value to 2 x values?

    RDKGames
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    (Original post by AliceAM342)
    Would the second mapping be a many-one relation, because there is a single y value to 4 x values.
    Yes it's many-one, but a single y only corresponds to two x values, not four.

    For the third mapping would it be a many- one relation, because there is a single y value to 2 x values?
    No. Think about what you're saying - can you give an example where two different x \geq 0 values output a single y value for y=\sqrt{x} ? More importantly, do you know how this function even looks like?
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    I’m sorry I am confused by the last mapping and cannot answer your question. I had been doing mappings, functions etc. But not with equations until now, only a series of numbers for the domain and range.

    I thought that there were 4 x values because the second mapping has x^4.

    For the third mapping I am lost, sorry 😯

    RDKGames
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    (Original post by AliceAM342)
    For the third mapping I am lost, sorry 😯

    RDKGames
    It's a one-one mapping. For each x \geq 0 you get a single output, and each y \geq 0 corresponds to a single input.
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    (Original post by RDKGames)
    It's a one-one mapping. For each x \geq 0 you get a single output, and each y \geq 0 corresponds to a single input.
    Thank you very much for your help, I really appreciate it. I understand (mainly) the third mapping.

    I will do some research to understand this topic more.
    Thank you 😊 RDKGames
 
 
 
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