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Math Help - Function Help

  1. #1
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    Function Help

    Decide whether the given relation defines y as a function of x. Give the domain...

    So, would you just put a number in y like 1 and keep incrementing and cubing the value to see if any x-values match or not?

    y=x^3

    What about something like this...

    y= Square root of 4x+2
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  2. #2
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    I would plug numbers into y then solve for x to find the domain
    <br />
y=x^3<br />
    As for,
    <br />
y=\sqrt{4x+2}<br />
    I would clear the radical from said equation entirely by squaring both sides. Then, I would start plugging numbers into y^2 and solve for x. You have to be careful though, and then check your results in the origional equation for any extraneous solutions. Give it a shot. Start with -2 on up to 2 and see what happens. Remember to check your solutions.
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  3. #3
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    That is
    <br />
(y)^2=(\sqrt{4x+2})^2<br />
    is equivalent to
    <br />
y^2=4x+2<br />
    this is because
    <br />
(\sqrt{4x+2})^2 = 4x+2 <br />
    and
    <br />
(y)^2 = y^2<br />
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  4. #4
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    Ok, thanks! I thought it was something like that, but the book doesn't explain it very well.
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  5. #5
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    Bam ba lam!!
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