1. Not finding help here? Sign up for a free 30min 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!

Help (Challenge proplems)

  1. Jun 22, 2007 #1

    I tried this for the first one but :confused:

    Attached Files:

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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar

    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


    User Avatar

    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


    User Avatar
    Science Advisor
    Homework Helper

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

Have something to add?

Similar Discussions: Help (Challenge proplems)
  1. Solving a log proplem! (Replies: 11)

  2. Integral challenge (Replies: 7)

  3. Challenge Problem (Replies: 1)