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Intermediate Value Theorem problem

  1. Sep 22, 2006 #1
    I have a problem that I am stuck with, if anyone please can give me any hint, that'd be great.

    f is a continuous function at the interval [a,b], and f(a) not euqal to f(b).
    q and p are two positive integers, and they do not equal zero.

    f : [a,b] ---> R

    show that :

    there is a number c that belongs to the interval [a,b] such as:
    pf(a) + qf(b) = (p+q)f(c)
    I only know that you should use the Intermediate Value Theorem, I only learned that this morning and I have no clue how to demonstrate this problem. Thanks for your help...
  2. jcsd
  3. Sep 22, 2006 #2


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    What does the intermediate value theorem say?
  4. Sep 22, 2006 #3
    " If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c. "
  5. Sep 22, 2006 #4
    make a little change and it will be much easier to prove it.

    let [tex]\frac{p}{p+q}=\lambda[/tex]
    because p and q are all positive, so [tex]\lambda[/tex] is some value at the interval (0,1)
    then we get [tex]\lambda f(a)+(1-\lambda)f(b)=f(c)[/tex]

    now you can use the Intermediate Value Theorem to prove it.
  6. Sep 22, 2006 #5
    I have to show an equality, I think that means I have to go from the first side, and reach the second part... and I still dont know how to prove your last line with the Intermediate Value Theorem
  7. Sep 22, 2006 #6
    supposing that the f value is at the interval [M,N]
    for [tex]\lambda[/tex] and [tex]1-\lambda[/tex] are positive,f(a) not eqal to f(b)
    [tex]\lambda f(a)+(1-\lambda)f(b) >\lambda M +(1-\lambda)M = M[/tex]

    [tex]\lambda f(a)+(1-\lambda)f(b) <\lambda N +(1-\lambda)N =N [/tex]
    so the value of [tex]\lambda f(a)+(1-\lambda)f(b)[/tex] is also at [M,N] according to the Intermediate Value Theorem,there exists some value at [a,b] that

    [tex]f(c)=\lambda f(a)+(1-\lambda)f(b)[/tex]
  8. Sep 22, 2006 #7
    I came out to this problem now;
    According to the IVT, knowing that f(a) not eqal to f(b), we have:
    f(a) < f(c) < f(b)
    so, f(a) is not equal to f(c), and, f(b) is not equal to f(c).
    pf(a) not equal to pf(c), and qf(b) is not equal to qf(c).
    if we add them together, we'll have:
    pf(a) + qf(b) is not equal to (p+q)f(c)
    and that doesn't not match with what I have to demonstrate.. ?!?!
  9. Sep 22, 2006 #8


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    But f(a)<(pf(a)+qf(b))/(p+q)<f(b). The IVT says any number satisfying this property has some c with f(c) equal to the number.
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