Calculus issues

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Lime
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Calculus issues

Postby Lime » Wed Feb 02, 2011 5:24 am UTC

Alright, I need help solving what is probably an easy problem. I need to rationalise the numerator of

((1/√x)-1)/(x-1)

I tried multiplying by ((1/√x)+1)/((1/√x)+1), but it just made a mess of the denominator.

MidsizeBlowfish
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Re: Calculus issues

Postby MidsizeBlowfish » Wed Feb 02, 2011 5:36 am UTC

try combining 1/sqrt(x)-1 into one fraction first.

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Lime
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Re: Calculus issues

Postby Lime » Wed Feb 02, 2011 5:38 am UTC

Wait, nevermind, I lost my train of thought. Derp.

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phlip
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Re: Calculus issues

Postby phlip » Wed Feb 02, 2011 5:44 am UTC

MidsizeBlowfish wrote:try combining 1/sqrt(x)-1 into one fraction first.

It comes out much nicer if you rationalise the denominator of the 1/sqrt(x) part first, then combine everything into one fraction.

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Qaanol
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Re: Calculus issues

Postby Qaanol » Wed Feb 02, 2011 6:07 am UTC

This is not a calculus problem.
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vilidice
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Re: Calculus issues

Postby vilidice » Wed Feb 02, 2011 8:34 pm UTC

Sadly I think more often than previous the first time people are encountering rationalizing the denominator is in calculus; what's even sadder is that after calculus, it's rarely ever used again in math classes.

What I like to do is get everything in the form of exponents first, so something like:
((2/(root(2x))-2)/(x+3) could be written as: (root(2)^2/root(2)root(x)+(iroot(2))^2)/(x+3), then simplified a bit to:
(root(2)/root(x)+i^2root(2))/(x+3) to (root(2)[(root(x)-root(2)]^-1)(x+3)^-1 = root(2)[root(x)(x+3)-root(2)(x+3)]^-1
= root(2)/[(x^3/2)+3root(x)-root(2)x-3root(2)] = 1/[x^3+(3root(x)/root(2))-x-3] = [x^3-x+3root(x)/root(2)-3]^-1 and I think that's as simplified as it gets, yours should look a bit nicer though, since more things will divide out evenly, mine was actually quite a mess, but I didn't want to use anything I thought you might get in an assignment.


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