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Math Help - Mean Value Theorem

  1. #1
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    Oct 2007
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    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

    |f(t)|<=2t^2/(1-t^2).

    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
    Joined
    Dec 2008
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    319
    First we note that when 0 \le x \le \frac{1}{2}, \log\,(1-x^2)\le 0 and therefore that

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

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

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

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

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

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

    If this is so, then we must have

    \begin{aligned}<br />
-\log\,(1-0^2)-\frac{8}{3}0^2 &= 0 \\<br />
-\log\,(1-k^2)-\frac{8}{3}k^2 &= 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0\le k\le \frac{1}{2}.<br />
\end{aligned}

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

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

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

    implies

    <br />
\begin{aligned}<br />
\frac{16}{3}-\frac{2}{1-x^2} &= 0 \\<br />
\frac{16}{3}(1-x^2)-2 &= 0 \\<br />
16(1-x^2)-6 &= 0 \\<br />
16(1-x^2) &= 6 \\<br />
1-x^2 &= \frac{3}{8} \\<br />
x &= \frac{1}{2}\sqrt{\frac{5}{2}} > \frac{1}{2},<br />
\end{aligned}

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