0. 3) It proves that if a function f' is bounded on an interval, then f is uniformly continuous on that interval. 4) It proves that if a function satisfies certain conditions at three points, then its second derivative must be positive at some point."> 0. 3) It proves that if a function f' is bounded on an interval, then f is uniformly continuous on that interval. 4) It proves that if a function satisfies certain conditions at three points, then its second derivative must be positive at some point.">
Math 171 Solutions To Homework Problems Spring 2005
Math 171 Solutions To Homework Problems Spring 2005
Math 171 Solutions To Homework Problems Spring 2005
Spring 2005
Section 4.3
cos x ex
1(b) Evaluate the limit if it exists: lim+
.
x0 log(1 + x2 )
Since lim+ cos x ex = lim+ log(1 + x2 ) = 0, we can use LHospitals Rule:
x0
x0
(sin x + ex )(1 + x2 )
cos x e
(cos x ex )0
sin x ex
=
lim
=
lim
=
lim
=
2x
log(1 + x2 )
2x
x0+
x0+ (log(1 + x2 ))0
x0+
x
lim
x0+
1+x2
1
4(a) Using (ex )0 = ex , (log x)0 = , and x = e log x , show that (x )0 = x1 for all
x
x > 0.
0
0
(x )0 = elog x
= e log x = e log x = x = x1
x
x
6 Let f be differentiable on a nonempty, open interval (a, b) with f 0 bounded on
(a, b). Prove that f is uniformly continuous on (a, b).
f 0 (c) =
. Therefore |f (x2 ) f (x1 )| = |f 0 (c)| |x2 x1 | < M = M
= .
x2 x 1
M
8 Let f be twice differentiable on (a, b) and let there be points x1 < x2 < x3 in (a, b)
such that f (x1 ) > f (x2 ) and f (x3 ) > f (x2 ). Prove that there is a point c (a, b) such
that f 00 (c) > 0.
Since x1 < x2 and f (x1 ) > f (x2 ), by the Mean Value Theorem there exists a point a (x1 , x2 )
f (x2 ) f (x1 )
such that f 0 (a) =
< 0.
x2 x 1
Since x2 < x3 and f (x3 ) > f (x2 ), by the Mean Value Theorem there exists a point b (x2 , x3 )
f (x3 ) f (x2 )
such that f 0 (b) =
> 0.
x3 x 2
Then by the Mean Value Theorem (applied to f 0 (x)) there exists a point c (a, b) such that
f 0 (b) f 0 (a)
f 00 (c) =
> 0.
ba
Section 4.4
1(b) Find all a R such that ax2 + 3x + 5 is strictly increasing on the interval (1, 2).
Let f (x) = ax2 + 3x + 5. Then f 0 (x) = 2ax + 3 is continuous. If f 0 (x) is negative at some
number, it is negative on some open interval, and then by the increasing/decreasing test, f (x)
is decreasing on that interval. Therefore if f (x) = ax2 + 3x + 5 is strictly increasing then
f 0 (x) 0 on (1, 2). So 2ax + 3 0 on (1, 2). Since f 0 (x) = 2ax + 3 is continuous, f 0 (2) 0.
3
So we have 4a + 3 0, so a .
4
3
Conversely, if a , then 2ax + 3 > 0 on (1, 2), and then f (x) is strictly increasing on
4
(1, 2).
1
f 0 (f 1 (2))
1
f 0 (0)
1
.
(b) (g 1 )0 (2).
(g 1 )0 (2) =
1
1
1
= 0
= .
g 0 (g 1 (2))
g (1)
e
(c) (f 1 g 1 )0 (2).
(f 1 g 1 )0 (2) = f 1 (2)(g 1 )0 (2) + (f 1 )0 (2)g 1 (2) = 0
1
1
1
+ 1= .
e
1
for x (, ).
1 + x2
1
for
1 x2
p
1 x2 .
1
1
=
.
2
tan (arctan x)
sec (arctan x)
0
1
.
1 + x2