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anonymous
 5 years ago
find the domain of the function
f(w)=w^33w+1/ (2w7)
anonymous
 5 years ago
find the domain of the function f(w)=w^33w+1/ (2w7)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2w7 \ne 0 \implies 2w \ne 7 \implies w \ne \frac{7}{2}\] \[\implies Domain: w\in (\infty,\frac{7}{2})\bigcup(\frac{7}{2},\infty)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please do explain to me how its done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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 2w7 is in the denominator to find what values for w are allowed we have to find the values where 2w7 is not equal to 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which is any value from negative infinity, up to (but not including) 7/2 then from greater than 7/2's on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but y equate 2w7 to 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not equating. I'm saying that 2w7 does NOT equal (\(\ne\)) 0. It's the same as solving for what w is if 2w7 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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