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Fifth exercise, in the solution, how did they get -6? In my eyes, logically, the outcome'd be non-existing, due to division by 0.
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This is calculus, that means the thing you should be interested in is the value of the function as x approachs 9, not when it equals 9 so the denominator will get very small but should not quite be 0.
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How do they get -6 then instead of 0? Or in this case, very close to -6 instead of very close to 0?
Also, how'd they get rid of the bottom part of the division in the next to one last exercise? |
OW MY BRAIN
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In the next to last question, they canceled the denominator with the expanded (x-4) cubed to get (x-4)|x-4| which tends towards 0.
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How does (x-4) cubed lead to 0
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Found what they did in both. Seems that they did this:
9-x = (sqrt(x)+3)(-sqrt(x)+3) And x-9 = (sqrt(x)+3)(sqrt(x)-3) So they can scratch (sqrt(x)+3), leaving (-sqrt(x)+3) <blablabla> / (sqrt(x)-3). Then they put the top - outside of the brackets, giving this: -(sqrt(x)-3) <blablabla> / (sqrt(x)-3) You can now scratch (sqrt(x)-3), leaving: -<blablabla>. In the 5th exercise, it ends up beautifully in -6. In the next-to-last one, it ends up in |x-4|? I haven't fully figured out why they tossed one out in front, so that it looks like (x-4)|x-4| Thanks goes to Pogo on another forum for pointing the sqrt(x) thingy out. |
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Meh, just a matter of not thinking "I don't understand this" on first sight, but try to follow the reasoning. That's how I'm improving on it, and I suck at math :ok:
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