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Thread: Prove a function decreases.

  1. #1
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    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?
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    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?
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    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 $\displaystyle f(x)=\frac{\log(x+1)}{\log(x)}$.

    Then prove that $\displaystyle f'(x)<0,~\forall x\in(1,\infty)$.
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    Re: Prove a function decreases.

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

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

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

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

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

    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?
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    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., $\displaystyle x\log(x) - (x+1)\log(x+1) < 0$. In other words we need to show that $\displaystyle x\log(x) < (x+1)\log(x+1)$.

    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: $\displaystyle a\ln b = \ln b^a$ and then try $\displaystyle e^{\mbox{LHS}}$ compared to $\displaystyle e^{\mbox{RHS}}$ where LHS = Left Hand Side and RHS = Right Hand Side. How does it alter the inequality?
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    Re: Prove a function decreases.

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

    Therefore, $\displaystyle x\log(x)<x\log(x+1)<(x+1)\log(x+1)$
    Last edited by Plato; Oct 1st 2013 at 10:53 AM.
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    Re: Prove a function decreases.

    Here is my complete proof, does this work:

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

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

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

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

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

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

    $\displaystyle x^x < (x+1)^{x+1}$

    Now $\displaystyle x^x < (x+1)^{x+1}$ is clearly true (or do I need to show something more) so it follows that the numerator of $\displaystyle f'(x)$ is negative $\displaystyle \forall x\in(1,\infty)$. Hence $\displaystyle f(x)$ is decreasing over the whole interval.
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    Re: Prove a function decreases.

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

    Edit:

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