My Math Forum Determine p(x) so that...

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October 1st, 2012, 10:00 PM   #1
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Determine p(x) so that...

Can someone tell me how to do this? Not just the answer, but like a step by step process on how to do it? IT'S DUE IN 2 HOURS AHHH!

I missed the day of class he went over this to study for another test... sigh
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 October 1st, 2012, 10:12 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Determine p(x) so that... We can find $p(x)$ by differentiating (using the quotient, power and chain rules): $\frac{d}{dx}$$\frac{x^2-4}{x^2+4}$$=\frac{$$x^2+4$$$$2x$$-$$x^2-4$$$$2x$$}{$$x^2+4$$^2}=\frac{16x}{$$x^2+4$$^2}$ $\frac{d}{dx}$$\frac{16x}{\(x^2+4$$^2}\)=\frac{$$x^ 2+4$$^2$$16$$-$$16x$$$$2\(x^2+4$$(2x)\)}{$$x^2+4$$^4}=\frac{16$$x^2+4$$$$\(x^2+4$$-4x^2\)}{$$x^2+4$$^4}=$ $\frac{16$$4-3x^2$$}{$$x^2+4$$^3}$ Hence: $\frac{d^2}{dx^2}$$\frac{1}{4}\(\frac{x^2-4}{x^2+4}$$\)=\frac{4$$4-3x^2$$}{$$x^2+4$$^3}$ Thus, we find: $p(x)=4$$4-3x^2$$=16-12x^2$
 October 1st, 2012, 10:27 PM #3 Member   Joined: Oct 2012 Posts: 39 Thanks: 0 Re: Determine p(x) so that... Thanks a bunch dude, Gotta write this in my notes!

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