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sjg13e
 one year ago
Another Partial Functions Problem
sjg13e
 one year ago
Another Partial Functions Problem

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sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0please give me second to upload the problem and my work

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x  14 }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{A}{x+7}+\frac{B}{x2}\] is a start

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x  14 }dx = \int\limits_{}^{}\frac{ 4 }{ (x1)(x+1) }dx = \frac{ A }{ x+7 } + \frac{ B }{ x2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you want to find A and B the snappy way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or the slower way (same thing really)

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets do it the real snappy way first

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0The way I did it: 4 = A(x2) + 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{4}{(x+7)(x2)}=\frac{A}{x+7}+\frac{B}{x2}\] to find \(A\) put your finger over the factor of \(x+7\) in \[\frac{4}{(x+7)(x2)}\]and replace \(x\) by \(7\) \[\frac{4}{\cancel{(x+7)}(72)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you get \[A=\frac{4}{9}\] instantly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you want to do it your "system of equations" long boring way and see if we can find the mistake?

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Then, \[\int\limits_{}^{}\frac{ 4 }{ x^2 + 5x  14 }dx = \int\limits_{}^{}\frac{ 4 }{ 5(x+7) }dx  \int\limits_{}^{}\frac{ 4 }{5(x2) }dx\] So the answer I got is \[= \frac{ 4 }{ 5 }\ln x+7  \frac{ 4 }{ 5 }\ln x2 + C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah the 5 is wrong, should be 9

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0I understand the method you used! Yes, can we try the boring way? Lol. I just want to make sure I can do both methods

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I don't know where I went wrong. Somehow I ended up with 5 instead of 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i like the "put your finger over the factor" method a lot more

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Haha, me too! But I need to be familiar with both methods for my exam

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok sure lets see the mistake \[A(x2)+B(x+7)=4\]is a start then \\[Ax2A+Bx+7B=4\]so \[A+B=0\\2A+7B=4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i see it right away since \(A=B\) you have \[2B+7B=4\\ B=\frac{4}{9}\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0I wrote 4 = 2A  7B instead of 4 = 2A + 7B.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh ok well glad we cleared that up!

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0then A = 4/9 and B = 4/9 and then we get the right answer haha. Thank you!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there 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

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0nice, maybe i'll come across the method again
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