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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 8 -- {\bf Due:}  Friday, April 19}
\bigskip
\noindent
{\it `A` Section}
\bigskip
\noindent
1.  For each of the following functions and intervals: First state whether 
the hypotheses of the MVT are satisfied for $f$.  Second: If they are satisfied, 
determine all $c \in (a,b)$ such that $f'(c) = {f(b) - f(a) \over b - a}$.  If
they are not satisfied, determine whether the conclusion of the MVT holds even so.
\bigskip
\item{(a)} $f(x) = {4x + 1\over x - 1}$ on $[a,b] = [0,2]$.  
\bigskip
\item{} {\it Solution:}  The equation $x - 1 = 0$ has a solution in $(0,2)$.  Hence the MVT does not apply, since $f$ is
not differentiable at every $x \in (0,2)$.  However the conclusion of the MVT
could still hold so we need to check.  We have $f(0) = -1$, $f(2) = 9$, so
$${f(2) - f(0) \over 2 - 0} = {10\over 2} = 5.$$
By the quotient rule, we have 
$$f'(x) = {(x - 1)(4) - (4x + 1)(1)\over (x - 1)^2} 
= {-5 \over (x - 1)^2}$$ 
This is always negative where it is defined, so there is no $c$ where $f'(c) = 5$.  
\bigskip
\item{(b)} $f(x) = x^{1/3}$ on $[a,b] = [0,1]$.
\bigskip
\item{} {\it Solution:}  $f$ is continuous on $[0,1]$ and differentiable on $(0,1)$
with $f'(x) = {1\over 3x^{2/3}}$.  Hence the MVT does apply.  We have
$${1\over 3 c^{2/3}} = {\sqrt{1} - \sqrt{0}\over 1 - 0} = 1$$
when $c^{2/3} = {1\over 3}$, or $c = {1\over 3\sqrt{3}}$.  
\bigskip
\item{(c)} $f(x) = \sin(\pi x)$ on $[0,3]$.
\bigskip
\item{} {\it Solution:}  Using problem 1 on the `B` section, $f$ is differentiable at all $x$, 
hence continuous on $[0,3]$.  The hypotheses of the MVT are satisfied.
By the derivative formula for $\sin(x)$ and the chain rule we have
$f'(x) = \pi\cos(\pi x)$.  Since $f(3) = 0$ and $f(0) = 0$, we are looking
for $c \in (0,3)$ where 
$$\pi \cos(\pi c) = 0,\ {\rm so}\ \cos(\pi c) = 0 $$
There are 3 such values of $c$ in the interval $(0,3)$, at $c = {1\over 2}, {3\over 2}, {5\over 2}$
\item{(d)} $f(x) = \cases{1/x & if $x > 0$ \cr
                           2 & if $x = 0$\cr}$ on $[a,b] = [0,2]$.  
\bigskip
\item{} {\it Solution:}  $f$ is not continuous at $0$ so the hypotheses
of the MVT are not satisfied.  ${f(2) - f(0) \over 2 - 0} = {-3\over 4}$.
For $x > 0$, $f'(x) = {-1\over x^2}$.  So there is a $c \in (0,2)$ where
$${-1\over c^2} = {-3\over 4},$$
namely $c = 2/\sqrt{3} \doteq 1.1547$.  Examples like this one show
that even when the hypotheses of the MVT are not satisfied, the conclusion
can still be true.  
\bigskip
\noindent
2.  Let $f(x) = x^3 - \lambda x^2 - \lambda^2 x$, where $\lambda > 0$ is
a constant.
\bigskip
\item{(a)}  Show that $f$ is a 1-1 function on the interval $I = [-\lambda/3,\lambda]$.
\bigskip
\item{} {\it Solution:}  We have 
$f'(x) = 3x^2 - 2 \lambda x - \lambda^2 = (3x + \lambda)(x - \lambda)$, which is strictly
negative on the open interval $(-\lambda/3, \lambda)$.  Hence by one of the corollaries
of the MVT (Theorem 4.3.15 in the text), $f$ is strictly decreasing 
on the interval $[-\lambda/3,\lambda]$.
\bigskip
\item{(b)}  Determine $f(I)$.
\bigskip
\item{} {\it Solution:}  $f(-\lambda/3) = 5\lambda^3/27 $ and $f(\lambda) = -\lambda^3$. 
So the image is $\left[-\lambda^3, {5\lambda^3\over 27}\right]$ using the Intermediate
Value Theorem.    
\bigskip 
\item{(c)}  Let $g$ be the inverse function of $f$ restricted to $I$.  What is
$g'(0)$?
\bigskip
\item{} {\it Solution:}  We have $g(f(x)) = x$ for all $x \in I$.  By the Chain 
Rule, $g'(f(x))f'(x) = 1$.  Since $f$ is 1-1 on $I$, there is just
one $x$ where $f(x) = 0$ for $x\in I$, namely $x = 0$.  Then $f'(0) = -\lambda^2$.  
Hence $g'(0) = {1\over f'(0)} = {-1\over \lambda^2}$.  
\bigskip
\noindent
3.  Let $f(x) = x^2 + x + 3$ on $[0,3]$.  
\bigskip
\item{(a)}  Compute $L(f,{\cal P})$ and $U(f,{\cal P})$ for 
the partition ${\cal P} = \{0,1,2,3\}$.
\bigskip
\item{} {\it Solution:}  We have $f'(x) = 2x + 1 > 0$ on $[0,3]$,
$f$ is strictly increasing on this interval.    
Therefore $m_1 = f(0) = 3$, $M_1 = f(1) = 5$, $m_2 = f(1) = 5$,
$M_2 = f(2) = 9$, $m_3 = f(2) = 9$ and $M_3 = f(3) = 15$.  Hence 
$$\eqalign{
L(f,{\cal P}) &= (3)(1) + (5)(1) + (9)(1) = 17\cr
U(f,{\cal P}) &= (5)(1) + (9)(1) + (15)(1) = 29\cr}$$
\item{(b)}  Compute $L(f,{\cal P})$ and $U(f,{\cal P})$ for 
the partition ${\cal P'} = \{0,1/2,1,2,5/2,3\}$ and verify that the
statement of Lemma 5.1.9 holds here. 
\bigskip
\item{} {\it Solution:}  For the partition ${\cal P'}$, which is a refinement of ${\cal P}$, 
we have 
$m_1 = 3$, $M_1 = f(1/2) = 15/4$, 
$m_2 = 15/4$, $M_2 = f(1) = 5$, 
$m_3 = 5$, $M_3 = f(2) = 9$, 
$m_4 = 9$, $M_4 = f(5/2) = 47/4$, 
$m_5 = 47/4$, $M_5 = 15$.
Hence 
$$\eqalign{
L(f,{\cal P'}) &= (3)(1/2) + (15/4)(1/2) + (5)(1) + (9)(1/2) + (47/4)(1/2) = 75/4\cr
U(f,{\cal P'}) &= (15/4)(1/2) + (5)(1/2) + (9)(1) + (47/4)(1/2) + (15)(1/2) = 99/4\cr}$$
As we expect from Lemma 5.1.9, since ${\cal P'}$ is a refinement of ${\cal P}$,
$$L(f,{\cal P}) \le L(f,{\cal P'}) \le U(f,{\cal P'}) \le U(f,{\cal P}).$$
\bigskip
\noindent
{\it `B` Section}
\bigskip
\noindent
1.  Recall that on a previous problem set, we showed
that 
$$\lim_{x \to 0} {\sin x \over x} = 1.\leqno(1)$$  
\item{(a)}  Show that 
$$\lim_{x \to 0} {1 - \cos(x)\over x} = 0.\leqno(2)$$  
You may only use the fact in equation (1) above and other general facts about limits. 
(In other words, no L'Hopital's Rule, which we have not discussed (yet) in this course -- see
part (e) of question 1 in the `B` section below.)
\bigskip
\item{} {\it Solution:}  We have 
$$\eqalign{
\lim_{x \to 0} {1 - \cos(x) \over x} &= \lim_{x \to 0} {1 - \cos(x) \over x}\cdot {1 + \cos(x)\over 1 + \cos(x)} \cr
&= \lim_{x \to 0} {1 - \cos^2(x) \over x(1 + \cos(x))}\cr
&= \lim_{x \to 0} {\sin^2(x) \over x (1 + \cos(x))}\cr
&= \lim_{x \to 0} {\sin(x) \over x} \cdot {\sin(x)\over (1 + \cos(x))}\cr
&= 1 \cdot 0 = 0\cr}$$
by the ``big theorem'' on limits.  
\item{(b)}  Use the limits in (1) and (2) above to show that if $f(x) = \sin(x)$
then $f$ is differentiable at all real $c$, and $f'(c) = \cos(c)$.  (Hint:
The best way to do this is to set up the limit for the derivative like this:
$$f'(c) = \lim_{x \to c} {\sin(x) - \sin(c)\over x - c} = 
\lim_{h\to 0} {\sin(c + h) - \sin(c)\over h}.$$
\item{} {\it Solution:}  We have 
$$\eqalign{\lim_{h\to 0} {\sin(c + h) - \sin(c)\over h} &=
\lim_{h\to 0} {\sin(c)\cos(h) + \cos(c)\sin(h) - \sin(c) \over h}\cr
&= \cos(c) \cdot \lim_{h \to 0} {\sin(h)\over h} 
- \sin(c) \cdot \lim_{h\to 0} {\cos(h) - 1\over h}\cr
&= \cos(c) \cdot 1 - \sin(c) \cdot 0\quad {\rm (using\ (1)\ and\ (2))}\cr
&= \cos(c).\cr}$$ 
\item{(c)}  Similarly, show that if $g(x) = \cos(x)$, then $g$ is differentiable
at all real $c$ and $g'(c) = -\sin(c)$. 
\bigskip
\item{} {\it Solution:}  We have
$$\eqalign{g'(c) = \lim_{x \to c} {\cos(x) - \cos(c)\over x - c} &= 
\lim_{h\to 0} {\cos(c + h) - \cos(c)\over h}\cr
&= \lim_{h\to 0} {\cos(c)\cos(h) - \sin(c)\sin(h) - \cos(c) \over h}\cr
&= -\sin(c) \cdot \lim_{h \to 0} {\sin(h)\over h} 
+ \cos(c) \cdot \lim_{h\to 0} {\cos(h) - 1\over h}\cr
&= -\sin(c) \cdot 1 + \cos(c) \cdot 0\cr
&= -\sin(c).\cr}$$  
\bigskip
\noindent
2.  (Applications and Extensions of the ``Most Valuable Theorem'')
\bigskip
\item{(a)}  Prove that if $f$ is differentiable on $(a,b)$ and there exists a 
positive constant $M$ such that $|f'(x)| \le M$ for
all $x \in (a,b)$, then $f$ is uniformly continuous on $(a,b)$.  
\bigskip
\item{} {\it Solution:}  By the MVT we have that if $x_1 < x_2$
are any elements of the interval $(a,b)$, then there is some $c$ in the 
interval $(x_1,x_2)$ such that 
$$f(x_2) - f(x_1) = f'(c)(x_2 - x_1).$$
Taking absolute values we obtain
$$|f(x_2) - f(x_1)| \le M |x_2 - x_1|.$$
This shows that $f$ is Lipschitz continuous on $(a,b)$ as in last week's
problem set, and it follows that $f$ is uniformly continuous on $(a,b)$.  
\bigskip
\item{(b)}  Let $f$ be a function such that the first $n$ derivatives
$f', f'', \ldots, f^(n)$ exist for 
all $x\in {\bf R}$, and assume that 
$$f(x_1) = f(x_2) = \cdots = f(x_{n+1})$$
for distinct $x_1,x_2, \ldots, x_{n+1} \in {\bf R}$.  Show that $f^{n}(c) = 0$
for some $c \in {\bf R}$.  (An induction proof is a possibility here.  What is the base case?)
\bigskip
\item{} {\it Solution:}  We prove this by induction on $n$.  The case $n = 1$ follows
from Rolle's Theorem applied to $f$ on the interval $[x_1,x_2]$.  
Since $f(x_1) = f(x_2)$, 
there is some $c \in (x_1,x_2)$ such that $f'(c) = 0$.  Now assume that whenever
$g$ is a function whose first $k$ derivatives exist on an interval containing
$k+1$ distinct points and $g(x_1) = g(x_2) = \cdots = g(x_{k+1)}$, then 
$g^{(k)}(c) = 0$ for some $c$ in that interval.  Now consider
a function $f$ whose first $k+1$ derivatives exist on some interval and 
assume $f(x_1) = f(x_2) = \cdots = f(x_{k+2})$ for distinct $x_1,\ldots,x_{k+2}$
in that interval.  We may assume here that 
$x_1 < x_2 < \cdots < x_{k+2}$ by relabeling if necessary.   Apply Rolle's Theorem
to $f$ on each of the intervals $[x_i,x_{i+1}]$ for $i = 1,2,\ldots, k+1$.  This
gives $c_i \in (x_i,x_{i+1})$ with $f'(c_1) = f'(c_2) = \cdots = f'(c_{k+1}) = 0$.  
Since those intervals are disjoint, it follows that the $c_i$ must be distinct.
Moreover, if the first $k+1$ derivatives of $f$ exist on an interval containing 
$x_1,\ldots, x_{k+2}$, then the first $k$ derivatives
of $f'$ exist on that same interval.  Hence we can apply the induction hypothesis
to $g = f'$. By the induction hypothesis, there exists some $c$ where the $k$th derivative of $f'$ is zero.  
But $(f')^{(k)} = f^{(k+1)}$ by the definition of the higher
derivatives of $f$.  
Therefore we have shown that there is a $c$ where $f^{(k+1)}(c) = 0$
and we are done by induction.  
\bigskip  
\item{(c)}  Let $f$ be differentiable and assume that $f'$ is strictly increasing
on ${\bf R}$.  If $f(a) = f(b)$, where $a < b$, then show that $f(x) < f(a) = f(b)$ for
all $a < x < b$.
\bigskip
\item{} {\it Solution:}  Let $a < x < b$.  The hypotheses of the MVT are satisfied
on both intervals $[a,x]$ and $[x,b]$.  Hence there are $c_1 \in (a,x)$
and $c_2 \in (x,b)$ such that $f'(c_1) = {f(x) - f(a)\over x - a}$ and 
$f'(c_2) = {f(b) - f(x) \over b - x}$.   If $f(x) \ge f(b) = f(a)$, then from these
consequences of the MVT, it would follow that $f'(c_1) \ge 0$ and $f'(c_2) \le 0$
But note that $a < c_1 < x < c_2 < b$, so by the 
assumptions, $f'(c_1) < f'(c_2)$.  Hence we cannot have $f(x) \ge f(a) = f(b)$.  
\bigskip
\item{(d)}  Let $f,g$ be continuous on $[a,b]$ and differentiable on $(a,b)$.  Show
there exists $c\in (a,b)$ such that $g'(c)(f(b) - f(a)) = f'(c)(g(b) - g(a))$.  
(Hint:  Consider the linear combination $h(x) = g(x)(f(b) - f(a)) - f(x)(g(b) - g(a))$. 
\bigskip
\item{} {\it Solution:}  $h$ is a linear combination of continuous functions on $[a,b]$, 
hence is continuous on $[a,b]$.  Similarly, it is differentiable on $(a,b)$.  We have 
$$h'(x) = g'(x)(f(b) - f(a)) - f'(x)(g(b) - g(a)).$$
Note that 
$$h(a) = g(a)f(b) - g(a)f(a) - f(a)g(b) + f(a)g(a) = g(a)f(b) - f(a)g(b).$$
Similarly
$$h(b) = g(b)f(b) - g(b)f(a) - f(b)g(b) + f(b)g(a) = g(a)f(b) - f(a)g(b).$$
By the MVT (in the Rolle's Theorem special case), there is a $c \in (a,b)$
where $h'(c) = 0$, and then
$$g'(c)(f(b) - f(a)) = f'(c)(g(b) - g(a)),$$
which is what we had to prove.   ({\it Comment:}  Applying the MVT separately to 
$f$ and $g$ on the interval $[a,b]$ and getting $f'(c) = {f(b) - f(a)\over b -a}$
for some $c \in (a,b)$ and $g'(c) = {g(b) - g(a) \over b - a}$, while tempting, 
{\it does not yield a correct proof} because there is no reason the two $c$'s
have to be the same(!))
\bigskip 
\item{(e)}  Use part (d) to deduce that if $f,g$ are differentiable on a deleted
neighborhood of $c$ with $\lim_{x\to c} f(x) = \lim_{x\to c} g(x) = 0$, if $g'(x) \ne 0$
on that deleted neighborhood and if
$$\lim_{x\to c} {f'(x)\over g'(x)} = L,\leqno(3)$$
then 
$$\lim_{x\to c} {f(x) \over g(x)} = L \leqno(4) $$
as well.  That is, show that the limit on the left side of the equation in (4) 
exists and equals $L$.  (This is a basic form of L'Hopital's Rule for limits.)
\bigskip
\item{} {\it Solution:}  Since $f$ and $g$ are differentiable and continuous on some open 
interval $(c - a,c)$ with $a > 0$ and $\lim_{x\to c^-} f(x) = \lim_{x \to c^-} g(x) = 0$,
if $f(c)$ and $g(c)$ are not defined directly, we can define $f(c) = g(c) = 0$
to get continuity at $c$.   We want to apply part (d) to $f,g$ on 
$[c - \delta,c]$, but we must make sure that $g(c-\delta) \neq 0$.  
If we did have $g(c - \delta) = 0$, then by Rolle's Theorem, 
$g'(\xi) = 0$ for some $\xi$ between $c - \delta$ and $c$.  
Since we assumed that $g'(x)$
was not zero on the whole deleted neighborhood, this cannot happen.  Hence for
each $\delta$,  we get that there is some $\xi \in (c - \delta,c)$ such that 
$${f'(\xi) \over g'(\xi)} = {f(c-\delta) - f(c)\over g(c-\delta) - g(c)} = {f(c-\delta) \over g(c-\delta)}.$$
Now let $\delta \to 0^+$ and use the assumption (3), we get (since $c - \delta < \xi < c$)
$$\lim_{\delta \to 0^+} {f(c-\delta) \over g(c - \delta)} = 
\lim_{\xi\to c^-} {f'(\xi)\over g'(\xi)} = L.$$
A similar argument ``from the other side'' shows
$$\lim_{\delta \to 0^+} {f(c+\delta) \over g(c + \delta)} = 
\lim_{\xi\to c^+} {f'(\xi)\over g'(\xi)} = L.$$
Therefore
$$\lim_{x\to c} {f(x)\over g(x)} = L$$
as required to establish (4).    ({\it Comment:}  If you assume more about 
$f,g$, there is also a somewhat simpler proof.  For instance, if you assume
that $f',g'$ exist and are continuous on a whole interval containing $c$ and
$g'(x) \ne 0$ on that interval, then you can argue like this
$$\eqalign{\lim_{x\to c} {f'(x)\over g'(x)}
&= {f'(c)\over g'(c)}\quad {\rm assuming\ } f',g'\ {\rm continuous\ at\ } c\cr
&=  {\lim_{x\to c}{f(x) - f(c)\over x - c} \over \lim_{x\to c}{g(x) - g(c) \over x - c}}\cr
&= \lim_{x\to c} {{f(x) - f(c)\over x - c} \over {g(x) - g(c) \over x - c}}\cr
&= \lim_{x \to c} {f(x) - f(c) \over g(x) - g(c)}\cr
&= \lim_{x\to c}{ f(x) \over g(x)} \quad \quad {\rm since\ } f(c) = g(c) = 0\cr}$$
I gave credit for this argument, but you should note that it uses a lot more
assumptions than what was given(!)
\bigskip
\noindent
3.  Show that $\int_0^3 f$ exists and determine its value if
$$f(x) = \cases{1 & if $0 \le x \le 1$\cr
                3 & if $1 < x \le 3$.\cr}$$
(That is, show that $f$ is integrable according to our definition and determine
the value of $\int_0^3 f$.)
\bigskip
\item{} {\it Solution:}  Let $\varepsilon > 0$ and consider the partition
$${\cal P}_\varepsilon = \left\{0, 1 - {\varepsilon\over 4}, 1 + {\varepsilon\over 4}, 3\right\}$$
For this partition we have 
$$m_1 = 1 = M_1, m_2 = 1\ {\rm while\ } M_2 = 3, m_3 = M_3 = 3.$$
Hence the difference 
$$U(f,{\cal P}_\varepsilon) - L(f,{\cal P}_\varepsilon)$$
is
$$
(1)\left(1 - {\varepsilon\over 4}\right) + (3)\left({\varepsilon\over 2}\right) + 
(3)\left(2 - {\varepsilon\over 4}\right) 
- (1)\left(1 - {\varepsilon\over 4}\right) - (1)\left({\varepsilon\over 2}\right) - (3)\left(2 - {\varepsilon\over 4}\right) 
= \varepsilon.$$
Since this can be made as small as we like, the function $f$ is integrable 
and the value of the integral is the limit as $\varepsilon \to 0$ of either
the upper or lower sum, namely
$$\int_0^3 f = 7.$$

\bye
