sjg13e
  • sjg13e
Partial Functions - Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sjg13e
  • sjg13e
please give me a second to upload the question and my work :)
anonymous
  • anonymous
it's alright
anonymous
  • anonymous
i wait ....

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sjg13e
  • sjg13e
\[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }\]
sjg13e
  • sjg13e
my 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
  • amistre64
apart from a typo or two its going fine so far
sjg13e
  • sjg13e
then \[x^2 - 3 = A(x+2) = B(x-2) \] \[x^2 - 3 = Ax + 2A + Bx - 2B\] \[x^2 - 3 = (A + B)x - 2A - 2B\]
sjg13e
  • sjg13e
so yes, sorry there are some typos!
amistre64
  • amistre64
dont bother distributing .. let x=2 and -2 to solve for A and B \[(2)^2 - 3 = A(2+2) + B(2-2)\] \[(-2)^2 - 3 = A(-2+2) + B(-2-2)\]
sjg13e
  • sjg13e
From 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
  • freckles
Partial most likely won't work until you do division. We need to divide first because you have an improper fraction.
amistre64
  • amistre64
... it works fine, just dont do the distribution part; that way we can zero out a term to solve
sjg13e
  • sjg13e
@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
  • amistre64
well, you saw why distribution cuases issues already :)
amistre64
  • amistre64
A(x-2) can be eliminated when x=2 B(x+2) can be eliminated when x=-2
amistre64
  • amistre64
(2)^2-3=A(2+2)+B(2-2) 4-3=A(4)+B(0) 1 = 4A ......................... (-2)^2-3=A(-2+2)+B(-2-2) 4-3=A(0)+B(-4) 1=-4B
freckles
  • freckles
Your 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
  • freckles
Anyways i will let Amistre help you because it is hard to read on this phone
amistre64
  • amistre64
x^2-3 + (1-1) = (x^2-4) + 1 if you really want to do the division first, but its not required
sjg13e
  • sjg13e
@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
  • amistre64
well, how else do you zero out a term? A(x+2) + B(x-2) 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
  • sjg13e
okay, that makes sense! thank you! i'll try that method and solve the problem again
amistre64
  • amistre64
\[\frac{x^2-3}{x^2-4}\implies \frac{(x^2-4)+1}{x^2-4}\implies1+\frac{1}{x^2-4}\] we still have to work the partial fraction even after the division, so the division is not necessary to me.
Jhannybean
  • Jhannybean
What i realized for most functions with polynomials in the denominator, it's either u-sub or partial fraction decomp. Hmm!
amistre64
  • amistre64
Jhanny, its a conspiracy :)
Jhannybean
  • Jhannybean
Knew it!
amistre64
  • amistre64
hmm, working it to a proper fraction might be needed ... never considered it before but the wolf suggests that it might be needed.
sjg13e
  • sjg13e
okay, so i would have to do long division before i decompose it?
freckles
  • freckles
yes @sjg13e as we notice \[\frac{A}{x-2}+\frac{B}{x+2}=\frac{(A+B)x+(2A-2B)}{x^2-4} \neq \frac{x^2-3}{x^2-4} \text{ for any real choice } \\ A \text{ and } B\] so division is definitely required before doing the partial fractions
sjg13e
  • sjg13e
Thank 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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