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anonymous

  • 5 years ago

find the domain of the function f(w)=w^3-3w+1/ (2w-7)

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  1. anonymous
    • 5 years ago
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    \[2w-7 \ne 0 \implies 2w \ne 7 \implies w \ne \frac{7}{2}\] \[\implies Domain: w\in (-\infty,\frac{7}{2})\bigcup(\frac{7}{2},\infty)\]

  2. anonymous
    • 5 years ago
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    please do explain to me how its done

  3. anonymous
    • 5 years ago
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    The domain of this function would be all values of w for which f is defined. Since a number divided by 0 is not defined, f is not defined when any denominator is 0. Since 2w-7 is in the denominator to find what values for w are allowed we have to find the values where 2w-7 is not equal to 0.

  4. anonymous
    • 5 years ago
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    Which is any value from negative infinity, up to (but not including) 7/2 then from greater than 7/2's on.

  5. anonymous
    • 5 years ago
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    but y equate 2w-7 to 0?

  6. anonymous
    • 5 years ago
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    I'm not equating. I'm saying that 2w-7 does NOT equal (\(\ne\)) 0. It's the same as solving for what w is if 2w-7 did equal 0, except that we are finding out what w isn't. That way we know what things we cannot have for w otherwise the denominator will be 0 and that cannot happen because dividing by 0 is not defined.

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