Partial Functions - Help!

- sjg13e

Partial Functions - Help!

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- schrodinger

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- sjg13e

please give me a second to upload the question and my work :)

- anonymous

it's alright

- anonymous

i wait ....

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## More answers

- sjg13e

\[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }\]

- 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

apart from a typo or two its going fine so far

- 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

so yes, sorry there are some typos!

- 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

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

Partial most likely won't work until you do division. We need to divide first because you have an improper fraction.

- amistre64

... it works fine, just dont do the distribution part; that way we can zero out a term to solve

- 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

well, you saw why distribution cuases issues already :)

- amistre64

A(x-2) can be eliminated when x=2
B(x+2) can be eliminated when x=-2

- 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

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

Anyways i will let Amistre help you because it is hard to read on this phone

- amistre64

x^2-3 + (1-1) = (x^2-4) + 1
if you really want to do the division first, but its not required

- 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

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

okay, that makes sense! thank you! i'll try that method and solve the problem again

- 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

What i realized for most functions with polynomials in the denominator, it's either u-sub or partial fraction decomp. Hmm!

- amistre64

Jhanny, its a conspiracy :)

- Jhannybean

Knew it!

- 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

okay, so i would have to do long division before i decompose it?

- 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

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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