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sjg13e

  • one year ago

Another Partial Functions Problem

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  1. sjg13e
    • one year ago
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    please give me second to upload the problem and my work

  2. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx\]

  3. anonymous
    • one year ago
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    \[\frac{A}{x+7}+\frac{B}{x-2}\] is a start

  4. sjg13e
    • one year ago
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    \[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ (x-1)(x+1) }dx = \frac{ A }{ x+7 } + \frac{ B }{ x-2 }\]

  5. anonymous
    • one year ago
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    you want to find A and B the snappy way?

  6. anonymous
    • one year ago
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    or the slower way (same thing really)

  7. sjg13e
    • one year ago
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    I was told to find A and B by doing system of equations. I do know that you can also find A and B quickly by letting x = -7 and x = 2, then solving for A and B. But let's do it the first way because that's where I messed up

  8. anonymous
    • one year ago
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    lets do it the real snappy way first

  9. sjg13e
    • one year ago
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    Okay

  10. sjg13e
    • one year ago
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    The way I did it: 4 = A(x-2) + B(x+7) 4 = Ax - 2A + Bx + 7B 4 = (A+B)x - 2A + 7B 4 = -2A - 7B A + B = 0 4 = -2(-B) - 7B A = -B 4 = 2B - 7B A = -(-4/5) 4 = -5B A = 4/5 -4/5 = B

  11. anonymous
    • one year ago
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    \[\frac{4}{(x+7)(x-2)}=\frac{A}{x+7}+\frac{B}{x-2}\] to find \(A\) put your finger over the factor of \(x+7\) in \[\frac{4}{(x+7)(x-2)}\]and replace \(x\) by \(-7\) \[\frac{4}{\cancel{(x+7)}(-7-2)}\]

  12. anonymous
    • one year ago
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    you get \[A=-\frac{4}{9}\] instantly

  13. anonymous
    • one year ago
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    you want to do it your "system of equations" long boring way and see if we can find the mistake?

  14. sjg13e
    • one year ago
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    Then, \[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x - 14 }dx = \int\limits_{}^{}\frac{ 4 }{ 5(x+7) }dx - \int\limits_{}^{}\frac{ 4 }{5(x-2) }dx\] So the answer I got is \[= \frac{ 4 }{ 5 }\ln |x+7| - \frac{ 4 }{ 5 }\ln |x-2| + C\]

  15. anonymous
    • one year ago
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    yeah the 5 is wrong, should be 9

  16. sjg13e
    • one year ago
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    I understand the method you used! Yes, can we try the boring way? Lol. I just want to make sure I can do both methods

  17. sjg13e
    • one year ago
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    Yeah, I don't know where I went wrong. Somehow I ended up with 5 instead of 9

  18. anonymous
    • one year ago
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    ok i like the "put your finger over the factor" method a lot more

  19. sjg13e
    • one year ago
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    Haha, me too! But I need to be familiar with both methods for my exam

  20. anonymous
    • one year ago
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    ok sure lets see the mistake \[A(x-2)+B(x+7)=4\]is a start then \\[Ax-2A+Bx+7B=4\]so \[A+B=0\\-2A+7B=4\]

  21. sjg13e
    • one year ago
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    Oh wait lol i see it

  22. anonymous
    • one year ago
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    oh i see it right away since \(A=-B\) you have \[2B+7B=4\\ B=\frac{4}{9}\]

  23. sjg13e
    • one year ago
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    I wrote 4 = -2A - 7B instead of 4 = -2A + 7B.

  24. anonymous
    • one year ago
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    ooh ok well glad we cleared that up!

  25. sjg13e
    • one year ago
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    then A = -4/9 and B = 4/9 and then we get the right answer haha. Thank you!

  26. anonymous
    • one year ago
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    yw

  27. anonymous
    • one year ago
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    there was a fellow here a while ago "eliasaab" i think, who had an amazingly snappy way to do this, even if you had annoying repeated roots i forget what it was though, had to do with derivatives

  28. sjg13e
    • one year ago
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    nice, maybe i'll come across the method again

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