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\begin{center}
{\bf \Large Math 112 Homework 7 Solutions}\medskip
\end{center}

{\bf Problem 1.}

Assume that $f'(x)>0$ for every $x\in (a,b)$, and consider two points
such that $x_1<x_2$ in $(a,b)$: by the mean value theorem, $f(x_2)-f(x_1)=
(x_2-x_1)f'(c)$ for some $c\in (x_1,x_2)$; since $f'(c)>0$ and $x_2-x_1>0$
we conclude that $f(x_1)<f(x_2)$, i.e.\ that $f$ is strictly increasing.
In particular $f$ is injective (one-to-one), and so there exists a
well-defined reciprocal function.

Let $I=f((a,b))$, and consider a point $y\in I$: by definition there exists
$x\in (a,b)$ such that $f(x)=y$, and since $g$ is the inverse function of
$f$ we have $g(y)=x$. We want to show that $g$ is differentiable at $y$ and
that $g'(y)=\frac{1}{f'(x)}$. For this purpose, we consider a sequence
$\{y_n\}$ of points of $I$ such that $y_n\neq y$ and $y_n\to y$, 
and we show that
$\frac{g(y_n)-g(y)}{y_n-y}$ converges to $\frac{1}{f'(x)}$. Since $y_n\in
I$, there exists $x_n\in(a,b)$ such that $f(x_n)=y_n$, and we have
$g(y_n)=x_n$, so that $\frac{g(y_n)-g(y)}{y_n-y}=\frac{x_n-x}{f(x_n)-f(x)}$.
In order to conclude by using the differentiability of $f$ at $x$, we first
need to show that $x_n\to x$ (i.e., that $g(y_n)\to g(y)$).

At least two methods can be used to show that $g$ is continuous at $y$:

1. Let $a',b'\in\R$ be such that $a<a'<x<b'<b$, and consider the restriction
$\tilde{f}$ of $f$ to the compact interval $[a',b']$. The function
$\tilde{f}$ is continuous and one-to-one (because it is strictly 
increasing). Therefore, by Theorem 4.17 the inverse mapping $\tilde{g}$
is also continuous. However $\tilde{g}$ is just the restriction of $g$
to the interval $f([a',b'])$, of which $y$ is an interior point (we have
$f(a')<f(x)=y<f(b')$ since $a'<x<b'$); therefore $g$ is continuous at $y$.

2. Observe that, if $y_1<y_2$, then $g(y_1)<g(y_2)$ (because if we had
$g(y_1)\ge g(y_2)$, then since $f$ is increasing we would obtain that
$y_1=f(g(y_1))\ge f(g(y_2))=y_2$, contradicting the assumption). Therefore
$g$ is also a strictly increasing function. Assume that $g$ is not
continuous at $y$: by Theorem 4.29 we have a simple discontinuity, 
i.e.\ $g(y-)<g(y+)$. Let $\alpha=g(y-)$ and $\beta=g(y+)$: we know that
$\forall t<y$, $g(t)\le\alpha$, so since $f$ is increasing we have
$t=f(g(t))\le f(\alpha)$ for every $t<y$; this implies that $f(\alpha)\ge y$.
Similarly, $\forall t>y$, $g(t)\ge\beta$, so $t=f(g(t))\ge f(\beta)$,
and therefore $f(\beta)\le y$. We conclude that $f(\alpha)\ge f(\beta)$
while $\alpha<\beta$, which contradicts the fact that $f$ is strictly
increasing. So $g$ is continuous at $y$.

Since $g$ is continuous at $y$ and $y_n\to y$, we can now conclude that 
$x_n=g(y_n)\to g(y)=x$. Therefore, by the definition of $f'(x)$, we have
$\frac{f(x_n)-f(x)}{x_n-x}\to f'(x)$, and therefore
$\frac{g(y_n)-g(y)}{y_n-y}=\frac{x_n-x}{f(x_n)-f(x)}\to \frac{1}{f'(x)}$.%
\medskip

{\bf Problem 2.}

Let $f:\R\to\R$ be the function
$f(x)=C_0x+\frac{C_1}{2}x^2+\dots+\frac{C_n}{n+1}x^{n+1}$.
Observe that $f(0)=0$ and $f(1)=0$ (this latter property follows from the
relation between the $C_i$). We know that $f$ is differentiable; therefore, 
by the mean value theorem there exists $x\in (0,1)$ such that
$f(1)-f(0)=(1-0)f'(x)$, i.e.\ $f'(x)=0$. Since $f'(x)=C_0+C_1x+\dots+C_n
x^n$, the conclusion follows.
\medskip

{\bf Problem 3.}

Fix $x_0\in [a,b]$, and let $M_0=\sup\,\{|f(x)|,\ x\in [a,x_0]\}$ and
$M_1=\sup\,\{|f'(x)|,\ x\in [a,x_0]\}$. The quantity $M_0$ is well-defined
because $f$ is continuous over the compact set $[a,x_0]$ and hence bounded,
and $M_1$ is also well-defined because $|f'(x)|\le A|f(x)|$ $\forall x$;
in fact we clearly have $M_1\le AM_0$. Pick a point $x\in [a,x_0]$: 
since $f(a)=0$, by the mean value theorem we have $f(x)=f(x)-f(a)=
(x-a)f'(c)$ for some $c\in (a,x)$; since $|f'(c)|\le M_1$ and 
$0\le x-a\le x_0-a$, we conclude that $|f(x)|\le (x_0-a) M_1\le (x_0-a)
AM_0$.

Since the bound $|f(x)|\le (x_0-a)AM_0$ holds for every $x\in [a,x_0]$, and
since $M_0=\sup\{|f(x)|,\ x\in [a,x_0]\}$, we conclude that $M_0\le
(x_0-a)AM_0$. Therefore, if $(x_0-a)A<1$ we must have $M_0=0$, which implies
that $f=0$ on the interval $[a,x_0]$. 

Choose elements $a=x_0<x_1<\dots<x_n=b$ such that $x_{i+1}-x_i<\frac{1}{A}$
for every value of $i$: for example, take $n$ large enough and set 
$x_i=a+\frac{i}{n}(b-a)$.
The above argument applied to $[a,x_1]\subset [a,b]$
shows that $f=0$ on $[a,x_1]$. We next consider the restriction of $f$ to
the smaller interval $[x_1,b]$: since we know from the previous step that 
$f(x_1)=0$, by applying the same argument to $[x_1,x_2]$ we show that $f=0$ 
on $[x_1,x_2]$. And so on by induction: once we have shown that $f(x_i)=0$,
we can consider the restriction of $f$ to $[x_i,b]$; by the above argument
$f=0$ on the interval $[x_i,x_{i+1}]$, and in particular $f(x_{i+1})=0$.
After $n$ steps, we conclude that $f=0$ over the entire interval $[a,b]$.
\medskip

{\bf Problem 4.}

Taylor's theorem (Theorem 5.15) with $\alpha=0$ and $\beta=1$ gives:
$$1=f(1)=f(0)+f'(0)+\frac12 f''(0)+\frac16 f^{(3)}(s)=\frac12 f''(0)+\frac16
f^{(3)}(s)\quad \text{for some $s\in(0,1)$.}$$
Applying again Taylor's theorem with $\alpha=0$ and $\beta=-1$ gives:
$$0=f(-1)=f(0)-f'(0)+\frac12 f''(0)-\frac16 f^{(3)}(t)=\frac12
f''(0)-\frac16 f^{(3)}(t)\quad \text{for some $t\in (-1,0)$.}$$
Subtracting the second identity from the first, we find that there exist
$s\in (0,1)$ and $t\in (-1,0)$ such that
$$1=\frac16 f^{(3)}(s)+\frac16 f^{(3)}(t),$$
i.e.\ $f^{(3)}(s)+f^{(3)}(t)=6$. We conclude that $f^{(3)}(s)\ge 3$
or $f^{(3)}(t)\ge 3$, which implies the desired conclusion.

\medskip

{\bf Problem 5.}

Note that the result is false if one does not assume $f$ to be continuous!
(see Chapter 6 Problem 1 for a counterexample).

We argue by contradiction: assume that there exists $x\in [a,b]$ such that
$f(x)>0$. Choose $0<\epsilon<\frac{1}{2}f(x)$: since $f$ is continuous at $x$,
there exists $\delta>0$ such that $\forall y\in [a,b]$, $|x-y|<\delta$
$\Rightarrow$ $|f(x)-f(y)|<\epsilon$. In particular, let
$\alpha=x-\frac{1}{2}\delta$ and $\beta=x+\frac{1}{2}\delta$ (if one of
these numbers lies outside of $[a,b]$ we set $\alpha=a$ or $\beta=b$
instead). We have $\alpha<\beta$, and $\forall y\in [\alpha,\beta]$,
$f(y)\ge f(x)-\epsilon>\frac{1}{2}f(x)$. Now consider the partition
$P=\{a,\alpha,\beta,b\}$ of $[a,b]$: since $f\ge 0$, the infimum of $f$
over each sub-interval is non-negative, and moreover $\inf_{[\alpha,\beta]}
f\ge \frac{1}{2}f(x)$. Therefore we have $L(P,f)\ge
0+\frac{1}{2}f(x)(\beta-\alpha)+0>0$. As a consequence,
$\int_a^b f\,dx=\underline\int_a^b f\,dx\ge L(P,f)>0$, which
is a contradiction. So $f(x)=0$ $\forall x\in[a,b]$.%

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