Results 1 to 9 of 9

Math Help - how to find the max,min,sup,inf of these cases..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    how to find the max,min,sup,inf of these cases..

    i made a limit on both infinity and minus infinity for them

    and i tried to find but its not working

    http://img201.imageshack.us/img201/5458/23597303em5.gif
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    The domain of the function \mathrm{ln}(x) is (0,\infty), with \lim_{x\to0}\mathrm{ln}(x)=-\infty.
    Let
    f(x):=\frac{1}{1+(\mathrm{ln}(x))^{2}},
    then
    f^{\prime}(x):=-\frac{2\mathrm{ln}(x)}{x(1+(\mathrm{ln}(x))^{2})^{  2}},
    which implies f^{\prime}(1)=0.
    Therefore its maximum value is f(1)=1.
    On the other hand, f attains its inferior value as x\to\infty (as you have shown).

    For the second one, let
    g(x):=x+\frac{1}{x}
    for x>0 (I believe it must be given on the positive halfline).
    Clearly, as x\to0 or x\to\infty, we have g(x)\to\infty (superior value).
    And on the other hand, we have
    g^{\prime}(x)=1-\frac{1}{x^{2}},
    which implies g^{\prime}(1)=0
    and thus g(x)\geq g(1)=2 for all x>0 (minimum value).

    You may similarly show the third one too, I have to say that this one is very interesting, just show that its maximum values decrease and similarly show that its minimum values increase.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    so the max and the sup of the first one is 1
    what is the min and inf of the first one?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    <br /> <br />
g(x):=x+\frac{1}{x}<br />

    the maximal number is infinty
    when x->+infinty

    the smallest value is found when
    x->-infinity

    ??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    regarding the third one:
    the derivative is
    (cosx * x - sin x)/x^2

    how to find the extreme points in order to find max min
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Quote Originally Posted by transgalactic View Post
    so the max and the sup of the first one is 1
    what is the min and inf of the first one?
    \lim_{x\to\infty}\ln(x)=\infty implies \lim_{t\to\infty}\frac{1}{1+(\ln(x))^{2}}=0 which is the inferior value, because there is no x_{0} point that \frac{1}{1+(\ln(x_{0}))^{2}}=0 holds.

    Quote Originally Posted by transgalactic View Post
    <br /> <br />
g(x):=x+\frac{1}{x}<br />

    the maximal number is infinty
    when x->+infinty

    the smallest value is found when
    x->-infinity

    ??
    As I said before x+\frac{1}{x} is unbounded; i.e., its superior value is \infty.
    And it takes its minimum value at x=1 with 1+\frac{1}{1}=2.

    Note. To find the extreme points of a function f, just find the zeros of the equation f^{\prime}=0.
    Say x_{0} is such a point that f^{\prime}(x_{0})=0, then if f^{\prime\prime}>0 we say that f has a minimum point at x_{0}, and if f^{\prime\prime}<0 we say that f has a maximum value at x_{0}.

    PS. I just now saw that it seems hard by proving in the way I told you before.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    in the third case the sinus function
    i get
    cosx * x -sinx =0

    how to solve this equation??
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by transgalactic View Post
    in the third case the sinus function
    i get
    cosx * x -sinx =0

    how to solve this equation??
    An obvious solution is x = 0. The function f(x) = \frac{\sin x}{x} has a removable singularity at x = 0. It makes sense to define f(0) = 1.

    Then the global maximum of f(x) = \frac{\sin x}{x} is located at (0, 1).

    The other local maxima (and local minima) cannot be found algebraically. You will need to use a numerical procedure to get approximate solutions.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Quote Originally Posted by transgalactic View Post
    in the third case the sinus function
    i get
    cosx * x -sinx =0

    how to solve this equation??
    Yes it is not easy, thus, we have to find other ways to find max and min values of this function.

    May the the following analysis may be helpful to you.
    Let f(t):=\sin(t)/t for t\in[0,\infty).
    Clearly, for all t\in[1,\infty), we have f(t)\leq|\sin(t)|\leq1.
    Thus, if there exists a point t_{0}\in[0,1] such that f(t_{0})\geq1 holds, then we can say that f(t_{0}) is the maximum value of f on [0,\infty).
    By calculating f^{\prime}(t)=\big(t\cos(t)-\sin(t)\big)/t^{2}, which implies f^{\prime}(0)=0, therefore f(0)=1\geq1 is its maximum value.

    I saw mr fantastic's arguments after posting the answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 18th 2011, 10:49 PM
  2. Question about cases
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 15th 2011, 05:57 PM
  3. CD Cases
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 13th 2008, 12:02 AM
  4. Ambiguous Cases
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 9th 2007, 03:23 PM
  5. Proof by cases
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 15th 2007, 03:06 AM

Search Tags


/mathhelpforum @mathhelpforum