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Homework Help: Help (Challenge proplems)

  1. Jun 22, 2007 #1

    I tried this for the first one but :confused:

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    Last edited: Jun 22, 2007
  2. jcsd
  3. Jun 22, 2007 #2


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    for the 1st question, let b=x, for the sake of tradition.

    find d/dx (e^x-x^e)

    find d/dx (e^x-x^e)=0 on the interval [0, infinity)

    prove that the zero was a minium, substitute the x it into y= e^x-x^e

    if the result is y=0, you have proved it

    thinking about the second question and the third question.
    Last edited: Jun 22, 2007
  4. Jun 22, 2007 #3


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    To find all the zeroes of that derivative, you probably essentially need to do the original problem anyway
  5. Jun 22, 2007 #4


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    nope, graph it and see
  6. Jun 22, 2007 #5
    1) For the first problem, look at the function f(x)=log(x)/x when x>=1.

    You can easily prove that it has a maximum at e. But then,


    whenever b>=1. This says b>=e*log(b). If you exponentiate this, you find


    2) For the second problem, look at the function f(x)=x^x. You can easily how that this has a minimum at 1/e. However, the limit as x goes to zero of x^x is 1. Furthermore, 1^1 is also 1. Thus x^x decreases between 0 and 1/e and increases between 1/e and 1. So if you have a number, say a, between 0 and 1/e, there must be a number between 1/e and 1, say b such that a^a=b^b.


  7. Jun 22, 2007 #6


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    Staff: Mentor

    Thread moved to Homework Help forums.
  8. Jun 23, 2007 #7
    this is the answer for the first 1
    we need to prove e^b>=b^e
    since ln(x) (natural lorarithm of x is continuous and monotonic increasing function we can take the natural logarithm of both sides of the inequality without disrupting it
    so we may prove ln(e^b)>=ln(b^e)
    by simplification
    so we only have to prove that b>=e*ln(b)
    it is readiely seen true for b=1 and b=e
    (for b=e) we have equality
    let's look at the two sides of the inequality as two functions of b
    the derivatives of the functions are
    for b>e df1(b)/db=1 and df2(b)/db=e/b<1
    so that from value of b>e f1 increases faster than f2 and since they both start at the same value at b=e the inequality holds for b>e
    for 1<b<e we have df2(b)/db=e/b>1 since b<e examining the regression of b from e to 1 it is regressing at a rate that is larger than 1 after starting from equal value of e it goes down faster than b therefore we have proven
  9. Jun 23, 2007 #8
    does anybody know how to solve the third one??:surprised
  10. Jun 23, 2007 #9


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    For the third one, I would beging with:

    P(x) = Ax^2 + Bx + C ; A,B,C are rational coefficients.

    Then I would a partial fraction decomposition:

    [tex] R(x) = \dfrac{Ax^{2} + Bx + C}{x^{3} (x-1)} = \dfrac{a_{0} ^{(1)}}{x^{3}} + \dfrac{a_{0} ^{(2)}}{x^{2}} + \dfrac{a_{0} ^{(3)}}{x} + \dfrac{a_{1} ^{(1)}}{x-1} [/tex]

    To find out the coefficients [tex] a_{1} ^{(1)} [/tex]....

    ; you just write

    [tex] R(x) = \dfrac{P(x)}{x^{3} (x-1)} [/tex]

    then you get a nice linear system of equations.

    After this, do the integration (remember the constant of integration), then see what values of [tex] a_{1} ^{(1)} [/tex] , and hence also values of A,B and C, that gives you a rational function.
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