1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
  2. jcsd
  3. Oct 3, 2009 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof: Solve Rational equation for interval
Loading...