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Thread: Mean Value Theorem

  1. #1
    Oct 2007

    Mean Value Theorem

    Let f(x)=log(1-x^2)

    Use the mean value theorem to show that |f(x)|<=(8x^2)/3 for 0<=x<=1/2.

    My attempt was this:

    f'(x)=-2x/(1-x^2), and so the is some c such that 0<=c<=t, and

    -2c/(1-c^2)=log(1-t^2)/t, so,

    |f(t)|=2ct/(1-c^2). But since 2c/(1-c^2) is increasing on 0<=c<=t, then


    Is this right? Am I going in the right direction? I'm not sure where to go from here.
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  2. #2
    Senior Member
    Dec 2008
    First we note that when $\displaystyle 0 \le x \le \frac{1}{2}$, $\displaystyle \log\,(1-x^2)\le 0$ and therefore that

    $\displaystyle |f(x)| = -f(x) = -\log\,(1-x^2).$

    We want to compare $\displaystyle |f(x)|$ and $\displaystyle g(x)=\frac{8}{3}x^2$ to show that $\displaystyle |f(x)|$ does not surpass $\displaystyle g(x)$ on $\displaystyle \left[0,\,\frac{1}{2}\right]$.

    To do this, we may use a proof by contradiction. It can be easily shown that there are $\displaystyle 0 \le x \le \frac{1}{2}$ for which $\displaystyle |f(x)| \le \frac{8}{3}x^2$. Suppose that for some $\displaystyle 0 \le h \le \frac{1}{2}$,

    $\displaystyle |f(h)| > \frac{8}{3}h^2.$

    By the Intermediate Value Theorem, there must be a number $\displaystyle k$ between $\displaystyle 0$ and $\displaystyle \frac{1}{2}$ such that

    $\displaystyle |f(k)| = \frac{8}{3}k^2$.

    If this is so, then we must have

    $\displaystyle \begin{aligned}
    -\log\,(1-0^2)-\frac{8}{3}0^2 &= 0 \\
    -\log\,(1-k^2)-\frac{8}{3}k^2 &= 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0\le k\le \frac{1}{2}.

    By the Mean Value Theorem, there must be a number $\displaystyle c$ between $\displaystyle 0$ and $\displaystyle k$ such that the derivative of $\displaystyle F(x)=\frac{8}{3}x^2-|f(x)|=\frac{8}{3}x^2+\log\,(1-x^2)$ at $\displaystyle c$, $\displaystyle F'(c)$, vanishes.

    However, solving $\displaystyle F'(x)$ for $\displaystyle 0$, we see that

    $\displaystyle \begin{aligned}
    F'(x) &= \frac{d}{dx}\left(\frac{8}{3}x^2+\log\,(1-x^2)\right) \\
    &= \frac{16}{3}x-\frac{2x}{1-x^2} = 0 \\


    \frac{16}{3}-\frac{2}{1-x^2} &= 0 \\
    \frac{16}{3}(1-x^2)-2 &= 0 \\
    16(1-x^2)-6 &= 0 \\
    16(1-x^2) &= 6 \\
    1-x^2 &= \frac{3}{8} \\
    x &= \frac{1}{2}\sqrt{\frac{5}{2}} > \frac{1}{2},

    provided that $\displaystyle x \ne 0$, contradicting the vanishing of $\displaystyle F'(c)$ where $\displaystyle c$ was supposed to lie between $\displaystyle 0$ and $\displaystyle \frac{1}{2}$.
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