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\noindent {\bf Question}
\noindent Use the mean value theorem to prove each of the
following statements.
\begin{enumerate}
\item If $g'(x)$ is a polynomial of degree $n-1$, then $g(x)$ is a
polynomial of degree $n$;
\item $x/(x+1)<\ln(1+x)0$;
\item $\sin(x)0$;
\end{enumerate}
\medskip
\noindent {\bf Answer}
\noindent \begin{enumerate}
\item Suppose that $g'(x) = a_{n-1} x^{n-1} + a_{n-2} x^{n-2}
+\cdots + a_1 x + a_0$, and consider the new function $h(x) =
\frac{1}{n} a_{n-1} x^n + \frac{1}{n-1} a_{n-2} x^{n-1} +\cdots +
\frac{1}{2} a_1 x^2 + a_0 x - g(x)$. Note that since $g$ and
polynomials are differentiable, and hence continuous, on all of
${\bf R}$, we have that $h$ is differentiable, and hence
continuous, on all of ${\bf R}$. Also, $h'(x) = a_{n-1} x^{n-1} +
a_{n-2} x^{n-2} +\cdots + a_1 x + a_0 - g'(x) = 0$ for all $x\in
{\bf R}$.
\medskip
\noindent For $x_0 >0$, apply the mean value theorem to $h$ on the
interval $[0, x_0]$. Since $h$ is continuous on $[0, x_0]$ and
differentiable on $(0, x_0)$, the mean value theorem yields that
there exists some $c$ in $(0, x_0)$ so that $h(x_0) - h(0) = h'(c)
(x_0 -0) = 0$, since $h'(c) =0$. That is, $h(x_0) =h(0)$ for all
$x_0 >0$. As above, we also get that $h(x_0) =h(0)$ for all $x_0
<0$ by applying the mean value theorem to $h$ on the interval
$[x_0, 0]$.
\medskip
\noindent Hence, setting $b =h(0)$, we have that $h(x) =b$ for all
$x\in {\bf R}$. Substituting in the definition of $h$, this
yields that $\frac{1}{n} a_{n-1} x^n + \frac{1}{n-1} a_{n-2}
x^{n-1} +\cdots + \frac{1}{2} a_1 x^2 + a_0 x - g(x) = b$ for all
$x\in {\bf R}$, that is, $g(x) = \frac{1}{n} a_{n-1} x^n +
\frac{1}{n-1} a_{n-2} x^{n-1} +\cdots + \frac{1}{2} a_1 x^2 + a_0
x - b$ for all $x\in {\bf R}$, and so $g$ is a polynomial of
degree $n$.
\item This is a slightly different sort of argument, and we break it
into two pieces, corresponding to the two inequalities.
\medskip
\noindent Set $h(x) = x -\ln(x+1)$, and note that $h$ is
differentiable, and hence continuous, on $(-1,\infty)$. The two
cases, of $-1 < x <0$ and of $x >0$, are handled in the same
fashion, and we write out the details only for the case $x >0$.
Apply the mean value theorem to $h$ on any closed interval in $[0,
\infty)$. Note that $h(0) =0 -\ln(1) =0$. If there were another
point $x_0> 0$ at which $h(x_0) =0$, then by applying either
Rolle's theorem or the mean value theorem to $h$ on the interval
$[0, x_0]$, there would exist a point $c$ in $(0,x_0)$ at which
$h'(c) =0$. However, $h'(c) = 1 -\frac{1}{c+1}$, which is
non-zero for $c\ne 0$. Hence, $h(x)\ne 0$ for all $x\in
(0,\infty)$. By the intermediate value theorem, this forces either
$h(x) >0$ for all $x >0$ or $h(x) <0$ for all $x >0$ (because if
there are points $a$ and $b$ in $(0,\infty)$ at which $h(a) >0$
and $h(b) <0$, then there is a point $c$ between $a$ and $b$ at
which $h(c) =0$). Since $h(1) = 1 -\ln(2) = 0.3069... >0$, we
have that $h(x) >0$ on $(0,\infty)$, that is, that $x > \ln(x+1)$
for all $x >0$, as desired. (As noted above, the argument to show
that $h(x) >0$ for $-1 \ln(x+1)$
for $-1 -1$. (As above, we give the details in the
case that $x >0$, and leave the case of $-1 0$ for $x >0$. In
particular, applying the mean value theorem to $g$ on the interval
$[0, x_0]$, we see that there is $c$ in $(0, x_0)$ so that $g(x_0)
-g(0) =g'(c) (x_0 -0) >0$, since both $g'(c) >0$ and $x_0 >0$.
Hence, $g(x_0) > g(0) = 0$ for all $x
>0$. That is, $\ln(x+1) > \frac{x}{x+1}$ for all $x >0$.
\item Here, set $g(x) = x -\sin(x)$. We wish to show that $g(x) >0$
for all $x >0$. First, note that since $-1 \le \sin(x) \le 1$ for
all $x\in {\bf R}$, we have that $g(x) >0$ for $x >1$, and so we
can restrict our attention henceforth to $0 1$ for $c\in
(0, 1)$, we have that $g(x_0) >0$ for all $0 0$ for all $x
>0$, as desired.
\end{enumerate}
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