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Math Help - domain help

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    domain help

    Hello,

    I am able to get all the other graphs on my paper along with their domain, range, zeros, symmetry with respect to origin, even/odd function, period and if it is a one-to-one function (for every y there is only one x). But I can not seem to figure this out. A graph along with the


    Domain:
    Range:
    Zeros:
    Symmetry with respect to origin (y-axis or origin):
    even/odd function:
    period (if it has one):
    is it a one-to-one function:

    would be great.

    this is the equation:

    f(x)=\sqrt{a^2-x^2}
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    y = \sqrt{a^2 - x^2}

    y^2 = a^2 - x^2<br />

    x^2 + y^2 = a^2

    does the last equation look familiar?

    how does the original equation relate to the above graphically?
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    reply

    wow, I can not believe I missed that, thank you. It is a circle. So the

    Domain: all real numbers
    Range: all real numbers
    Zeros: at \sqrt{a},0
    Symmetry with respect to origin (y-axis or origin): both
    even/odd function: both
    period (if it has one):none
    is it a one-to-one function:no

    correct?
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post

    correct?
    Nop,

    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post

    Range: all real numbers
    Square roots are not negative.
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    I was saying that range is all real numbers because it can be anywhere on the number line.
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  6. #6
    Math Engineering Student
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    Short counterexample: consider f(x)=\sqrt{4-x^2}, \exists\,x\in\text{Dom }f\,\Big|\,f(x)=-1 ?
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  7. #7
    MHF Contributor
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    actually, y = \sqrt{a^2 - x^2} is the upper semicircle.

    look at its graph.

    domain is -a < x < a

    range is 0 < x < a
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  8. #8
    Member OnMyWayToBeAMathProffesor's Avatar
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    aaa, I see, that makes a big difference. I understand 'skeeter' but 'Krizalid', I am sorry, I did not understand your symbols. Thank you
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