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sjg13e
 one year ago
Partial Functions  Help!
sjg13e
 one year ago
Partial Functions  Help!

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

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ x^2  3 }{ x^2  4 }\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0my work (it's wrong obviously lol): \[(x^2  4) = (x  2)(x +2)\] so \[\int\limits_{}^{}\frac{ x^2  3 }{ x^2  4 }= \frac{ x^2  3 }{ (x^2  2)(x^2 +2) } = \frac{ A }{ x 2 }+\frac{ B }{ x + 2 }\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1apart from a typo or two its going fine so far

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0then \[x^2  3 = A(x+2) = B(x2) \] \[x^2  3 = Ax + 2A + Bx  2B\] \[x^2  3 = (A + B)x  2A  2B\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0so yes, sorry there are some typos!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1dont bother distributing .. let x=2 and 2 to solve for A and B \[(2)^2  3 = A(2+2) + B(22)\] \[(2)^2  3 = A(2+2) + B(22)\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0From there, I tried to find the values of A and B A + B = 0 A = B \[2A  2B = 3 \] \[2A = 3 + 2(A)\] \[2A = 3  2A\] 0 = 3 (??)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Partial most likely won't work until you do division. We need to divide first because you have an improper fraction.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1... it works fine, just dont do the distribution part; that way we can zero out a term to solve

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 Wait, can you explain a little more why I shouldn't bother distributing and should instead just use substitution. @freckles, How did you know that I need to divide first? That step didn't immediately pop out to me

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, you saw why distribution cuases issues already :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1A(x2) can be eliminated when x=2 B(x+2) can be eliminated when x=2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1(2)^23=A(2+2)+B(22) 43=A(4)+B(0) 1 = 4A ......................... (2)^23=A(2+2)+B(22) 43=A(0)+B(4) 1=4B

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Your top poly and bottom poly had equal degree. Whenever the top poly has equal or greater degree to that of the bottom poly. I think division then partial

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Anyways i will let Amistre help you because it is hard to read on this phone

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1x^23 + (11) = (x^24) + 1 if you really want to do the division first, but its not required

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 okay, i understand the following work, but how did you know to let x = 2? in my calculus class we haven't really discussed this method within partial functions and thank you @freckles for your help! i'll try doing division first

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, how else do you zero out a term? A(x+2) + B(x2) anything times 0 is 0 ... so by observation letting x=2 or 2 gets rid of a term so that we can solve for the other.

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0okay, that makes sense! thank you! i'll try that method and solve the problem again

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{x^23}{x^24}\implies \frac{(x^24)+1}{x^24}\implies1+\frac{1}{x^24}\] we still have to work the partial fraction even after the division, so the division is not necessary to me.

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0What i realized for most functions with polynomials in the denominator, it's either usub or partial fraction decomp. Hmm!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1Jhanny, its a conspiracy :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1hmm, working it to a proper fraction might be needed ... never considered it before but the wolf suggests that it might be needed.

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0okay, so i would have to do long division before i decompose it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yes @sjg13e as we notice \[\frac{A}{x2}+\frac{B}{x+2}=\frac{(A+B)x+(2A2B)}{x^24} \neq \frac{x^23}{x^24} \text{ for any real choice } \\ A \text{ and } B\] so division is definitely required before doing the partial fractions

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @freckles! I solved the problem and read through some calculus notes and realized that it's required to get the leading x degree in the numerator to be less than the leading x degree in the denominator, and for this problem, it required long division
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