# Math 414 Professor Lieberman April 22, 2003 EXAM #3 SOLUTIONS ```Math 414
Professor Lieberman
April 22, 2003
EXAM #3 SOLUTIONS
1. By freshman calculus,
0
fr,s
(x) = rxr−1 sin(x−s ) − sxr−s−1 cos(x−s )
for x 6= 0. In addition,
hr sin(h−s )
= lim hr−1 sin(h−s ),
h→0
h→0
h
0
fr,s
(0) = lim
0 (0) exists for r &gt; 1.
so fr,s
0 is continuous at zero if r &gt; 1 and r &gt; s + 1, for example, r = 3 and s = 1.
(a) Therefore fr,s
0
0 (0) exists if r &gt; 1 and r ≤ s + 1, for example, r = 3
(b) fr,s is discontinuous at 0 but fr,s
and s = 2.
2. Abbreviate f (x) = x/(1 + x2 ), g(x) = arctan x and h(x) = x. Then
f 0 (x) =
1
2x2
1
−
, g 0 (x) =
, h0 (x) = 1,
2
2
2
1+x
(1 + x )
1 + x2
so f 0 (x) &lt; g 0 (x) &lt; h0 (x) if x &gt; 0, so f − g and g − h are strictly decreasing. Because
f (0) = g(0) = h(0), it follows that f (x) &lt; g(x) &lt; h(x) for x &gt; 0.
3. The easiest such function is
(
0
f (x) =
1
if x 6= 12 ,
if x = 12 .
It isn’t monotone because f takes on its maximum at an interior point and it is clearly
discontinuous at x = 1/2. To see that it’s integrable, note that the lower sum for any
partition is 0 and we can make the upper sum less than a given ε &gt; 0 by taking the partition
P = {x0 , x1 , x2 , x3 } with x0 = 1, x1 = max{1/4, (1 − ε)/2}, x2 = min{3/4, (1 + ε)/2}.
Another good example is Dirichlet’s function:
(
1
if x = pq in lowest terms,
f (x) = q
0 otherwise.
This was shown to be integrable in a homework problem.
4. From the definition,
f (x) − f (t)
.
t→x
x−t
f 0 (x) = lim
By hypothesis,
f (x) − f (t)
≤ |x − t|
x−t
and limt→x −|x − t| = limt→x |x − t| = 0, so the squeeze theorem implies that f 0 (x) = 0 for
all x. Therefore f is constant.
−|x − t| ≤
1
2
5. Actually, there is no such function, but here is a solution that would get full credit: If
f 0 (x) = 1/f (x), then f 0 (x)f (x) − 1 = 0, and f 0 (x)f (x) − 1 is the derivative of f (x)2 /2 − x,
so f (x)2 /2 − x must be constant. Therefore
√
f (x) = 2x + C
for some constant C.
6. Method 1: From the chain rule,
sin(ln x)
d
(− cos(ln x)) =
.
dx
x
Because these functions are both continuous on [1, 2], the fundamental theorem of calculus
says that
Z 2
2
sin(ln x)
dx = − cos(ln x) = 1 − cos(ln 2).
x
1
1
Method 2: Set f (x) = sin x and g(x) = ln x. Then g is differentiable and g 0 is Riemann
integrable on [1, 2] and g([1, 2]) = [0, ln 2]. Also f is continuous on [0, ln 2], so the Change
of Variable Theorem says that
Z 2
Z ln 2
ln 2
sin(ln x)
dx =
sin x dx = − cos x = 1 − cos(ln 2).
x
0
1
0
```