1. ## 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.

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.

2. 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}
-\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}.
\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}
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 \\
\end{aligned}

implies


\begin{aligned}
\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},
\end{aligned}

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