Results 1 to 8 of 8
Like Tree3Thanks
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By SlipEternal

Math Help - Prove a function decreases.

  1. #1
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Prove a function decreases.

    Prove that the function f(x) = logx(x+1) decreases on the interval (1,+infinity).

    How should I approach this proof? Thoughts?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member beebe's Avatar
    Joined
    Aug 2011
    Posts
    54

    Re: Prove a function decreases.

    In general, how do you know that a function increases or decreases? Take an easier example, I know that the function g(x)=x increases over its whole domain. How could I prove that to you?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Prove a function decreases.

    Quote Originally Posted by vidomagru View Post
    Prove that the function f(x) = logx(x+1) decreases on the interval (1,+infinity).
    How should I approach this proof? Thoughts?
    Write the function as f(x)=\frac{\log(x+1)}{\log(x)}.

    Then prove that f'(x)<0,~\forall x\in(1,\infty).
    Thanks from vidomagru
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Prove a function decreases.

    Quote Originally Posted by Plato View Post
    Write the function as f(x)=\frac{\log(x+1)}{\log(x)}.

    Then prove that f'(x)<0,~\forall x\in(1,\infty).
    So \frac{d}{dx}f(x)=\frac{x\log(x) - (x+1)\log(x+1)}{x(x+1)\log(x)\log(x)}.

    Since we need to show that f'(x)<0,~\forall x\in(1,\infty), I guess the next step would be to consider the numerator and denominator separately. Consider the denominator first:

    x(x+1)\log(x)\log(x) is always positive since x, x+1, \log(x) are by definition positive over the interval (1,\infty).

    It remains to show that the numerator is negative over the whole interval, i.e., x\log(x) - (x+1)\log(x+1) < 0. In other words we need to show that x\log(x) < (x+1)\log(x+1).

    I am not sure how to actually show this, since it is so obvious to me by looking at the graph. I thought about trying some kind of induction but we are not operating over N so it would be hard. Any further guidance you can give me?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,794
    Thanks
    694

    Re: Prove a function decreases.

    Quote Originally Posted by vidomagru View Post
    It remains to show that the numerator is negative over the whole interval, i.e., x\log(x) - (x+1)\log(x+1) < 0. In other words we need to show that x\log(x) < (x+1)\log(x+1).

    I am not sure how to actually show this, since it is so obvious to me by looking at the graph. I thought about trying some kind of induction but we are not operating over N so it would be hard. Any further guidance you can give me?
    You are on the right track. Use the rules you know about logarithms: a\ln b = \ln b^a and then try e^{\mbox{LHS}} compared to e^{\mbox{RHS}} where LHS = Left Hand Side and RHS = Right Hand Side. How does it alter the inequality?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Prove a function decreases.

    Quote Originally Posted by vidomagru View Post
    Since we need to show that f'(x)<0,~\forall x\in(1,\infty), I guess the next step would be to consider the numerator and denominator separately. Consider the denominator first:
    x(x+1)\log(x)\log(x) is always positive since x, x+1, \log(x) are by definition positive over the interval (1,\infty).
    If x>1 then \log(x)<\log(x+1) it is increasing.

    Therefore, x\log(x)<x\log(x+1)<(x+1)\log(x+1)
    Last edited by Plato; October 1st 2013 at 10:53 AM.
    Thanks from vidomagru
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Prove a function decreases.

    Here is my complete proof, does this work:

    So \frac{d}{dx}f(x)=\frac{x\log(x) - (x+1)\log(x+1)}{x(x+1)\log(x)\log(x)}.

    Since we need to show that f'(x)<0,~\forall x\in(1,\infty), I guess the next step would be to consider the numerator and denominator separately. Consider the denominator first:

    x(x+1)\log(x)\log(x) is always positive since x, x+1, \log(x) are by definition positive over the interval (1,\infty).

    It remains to show that the numerator is negative over the whole interval, i.e., x\log(x) - (x+1)\log(x+1) < 0. In other words we need to show that x\log(x) < (x+1)\log(x+1).

    x\log(x) < (x+1)\log(x+1)
    \log(x)^x < \log(x+1)^{x+1}
    e^{\log(x)^x} < e^{\log(x+1)^{x+1}}

    Note: e^{\log(x)} = x, so we have:

    x^x < (x+1)^{x+1}

    Now x^x < (x+1)^{x+1} is clearly true (or do I need to show something more) so it follows that the numerator of f'(x) is negative \forall x\in(1,\infty). Hence f(x) is decreasing over the whole interval.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,794
    Thanks
    694

    Re: Prove a function decreases.

    Quote Originally Posted by vidomagru View Post
    Now x^x < (x+1)^{x+1} is clearly true (or do I need to show something more) so it follows that the numerator of f'(x) is negative \forall x\in(1,\infty). Hence f(x) is decreasing over the whole interval.
    That is up to you. You can use the fact that for any 1<x, x^x = 1\cdot x^x < x\cdot x^x = x^{x+1} < (x+1)^{x+1} since x<x+1.

    Edit:

    If you want to use what Plato wrote, you would want to show that \ln(x) is a strictly increasing function on (1,\infty). That is obvious since (\ln(x))^\prime = \dfrac{1}{x} > 0 for any positive x. Hence, since x<x+1, \ln(x)<\ln(x+1), and the rest of what Plato wrote follows.
    Last edited by SlipEternal; October 1st 2013 at 11:10 AM.
    Thanks from vidomagru
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 26th 2010, 04:19 AM
  2. How do I know this series decreases or not?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 2nd 2010, 03:44 AM
  3. f decreases most rapidly at Q
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 2nd 2008, 01:21 AM
  4. Replies: 2
    Last Post: November 8th 2007, 03:45 PM
  5. graph that increases or decreases
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 9th 2006, 12:33 AM

Search Tags


/mathhelpforum @mathhelpforum