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Proof: Solve Rational equation for interval

  1. Oct 3, 2009 #1
    Prove that if c and d are positive, then the equation [tex] \frac{c}{x-2} + \frac{d}{x-4} = 0 [/tex] has at least one solution in the interval (2,4).
    I try to use the following theorem: If f is continuous on [a,b] and if f(a) and f(b) are nonzero and have opposite signs, then there is at least one solution of the equation f(x) = 0 in the interval (a,b).
    As the theorem requires f to be continuous on the closed interval [2,4], the open interval (2,4) is what is given, indeed there is a discontinuity at x=2 and at x =4. The equation can be rewritten as [tex] \frac{x-2}{x-4} + \frac{c}{d}=0 [/tex] x not equal to 2. I do not know of any other theorem to use. Can someone point me in the correct direction, please. Thank you.
     
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  3. Oct 3, 2009 #2

    LCKurtz

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    Look at the limits as x --> 2+ and x --> 4-. What does that tell you?
     
  4. Oct 3, 2009 #3
    [tex] \lim_{x\rightarrow 2^+} \mbox{ \lim_{x \rightarrow4^-} \mbox{ are } -\infty \mbox{ and } +\infty [/tex] I think or maybe the other way round, I have to check their signs using test points. It tells me that f(a)= -infinity and f(b)=+infinity
     
  5. Oct 3, 2009 #4

    LCKurtz

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    So can you get f to have opposite signs on the interval?
     
  6. Oct 3, 2009 #5
    Yes, I can get f to have opposite signs on the interval.And the conclusion of theorem follows
     
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