anonymous 5 years ago find the domain of the function f(w)=w^3-3w+1/ (2w-7)

1. anonymous

$2w-7 \ne 0 \implies 2w \ne 7 \implies w \ne \frac{7}{2}$ $\implies Domain: w\in (-\infty,\frac{7}{2})\bigcup(\frac{7}{2},\infty)$

2. anonymous

please do explain to me how its done

3. anonymous

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

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

5. anonymous

but y equate 2w-7 to 0?

6. anonymous

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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