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anonymous

  • 4 years ago

find the domain and range g(z)= 1 over square root 4-z square

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  1. lgbasallote
    • 4 years ago
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    find the value of z that will make the equation undefined... sq root (4-z^2) = 0 4 - z^2 = 0 4 = z^2 z = +/- 2 so...your domain is all values of x except for 2 and -2...which can be written as...(-inf, -2)U(2, inf) as for the range...cross multiply. y(sq root 4-z^2) = 1 4 - z^2 = 1/y^2 z^2 = 4 - 1/y^2 z^2 = (4y^2 - 1 )/y^2 get the sq root z = sq root (2y +1)(2y-1)/y anyways...your range is everything below and equal to zero...so your range is (-inf, 0)U(0, inf)

  2. anonymous
    • 4 years ago
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    how you throw away the square root to find range?

  3. lgbasallote
    • 4 years ago
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    y(sq root 4-z^2) = 1 divide both sides by y sq root 4-z^2 = 1/y square both sides... 4-z^2 = 1/y^2

  4. anonymous
    • 4 years ago
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    hmmm \[\frac{1}{\sqrt{4-z^2}}\] need \[4-z^2>0\] \[(2-z)(2+z)>0\] i think this is \[(-2,2)\] as the domain

  5. anonymous
    • 4 years ago
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    it is a square root right?

  6. anonymous
    • 4 years ago
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    is z a real variable?

  7. anonymous
    • 4 years ago
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    ooooooooh range i see ok that would be \[(\frac{1}{2},\infty)\]

  8. anonymous
    • 4 years ago
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    reason as follows. this beast can get as large as possible by making the denominator small, that is make x close to 2 or -2 the denominator is largest when x is 0, and when x is 0 you get \[\frac{1}{\sqrt{4}}=\frac{1}{2}\] so that is the minimum value of the function

  9. anonymous
    • 4 years ago
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    so to be more precise, range is \[[\frac{1}{2},\infty)\] or \[\frac{1}{2}\leq y<\infty\]

  10. anonymous
    • 4 years ago
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    actly which answer the correct one??? im blur

  11. anonymous
    • 4 years ago
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    \[g(z)=\frac{1}{\sqrt{4-z^2}}\] Domain \[(-2,2)\] Range \[[\frac{1}{2},\infty)\]

  12. anonymous
    • 4 years ago
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    r u sure????

  13. anonymous
    • 4 years ago
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    why u say that x>0

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