Math 3210, Fall 2013
Instructor: Thomas Goller
30 October 2013
Learning Celebration #3
Example (10 points)
Give an example of each of the following or state that no example exists:
(1) A function f defined on [1, 2] that does not take on every value between f (1) and f (2).
(2 points)
(
0 if x 2 [1, 32 ]
f (x) =
1 if x 2 ( 32 , 2]
(2) A function f satisfying limx!0+ f (x) =
(
1
if x < 0
x
f (x) =
1 if x > 0
1 and limx!0 f (x) = 1. (2 points)
(3) A function that is continuous on its domain but not uniformly continuous on its domain.
(2 points)
f (x) = x2
(4) A function that is uniformly continuous on an open interval I but not di↵erentiable on
I. (2 points)
f (x) = |x|
(5) A function defined on a closed bounded interval I that does not attain a maximum value
on I. (2 points)
(
x if x 2 [0, 1)
f (x) =
0 if x = 1
Computation (4 points)
sin x
cos x
and lim
x!0 x
x!0
x
(6) Compute lim
1
. Show your work! (4 points)
sin x
cos x
= lim
= cos 0 = 1.
x!0
x
1
cos x 1
sin x
lim
= lim
= sin 0 = 0.
x!0
x!0
x
1
lim
x!0
Math 3210, Fall 2013
Instructor: Thomas Goller
30 October 2013
Precision (6 points)
(7) State the Intermediate Value Theorem. (3 points)
If f is continuous on [a, b] and y 2 R is between f (a) and f (b), then there is c 2 [a, b]
such that f (c) = y.
(8) State the Mean Value Theorem. (3 points)
If f is continuous on [a, b] and di↵erentiable on (a, b), then there is c 2 (a, b) such that
f 0 (c) =
f (b)
b
f (a)
.
a
Proof (20 points)
(9) Let f (x) = 4x. Guess the limit limx!1 f (x), write your guess as a claim, and prove
the claim using the definition of limit. (5 points)
Claim. Let f (x) =
1.
4x. Then limx!1 f (x) =
Proof. Given M 2 R, let m =
M
.
4
Then for all x 2 R satisfying x > m,
f (x) =
4x <
4m = M.
(10) Let f (x) = 3x2 10. Guess the derivative of f , write your guess as a claim, and prove
the claim using the definition of the derivative. (5 points)
Claim. Let f (x) = 3x2
10. Then f 0 (x) = 6x.
Proof.
f (x + h)
h!0
h
f 0 (x) = lim
f (x)
= lim
3(x + h)2
h!0
10
h
(3x2
10)
6xh + 3h2
= lim
= lim (6x + 3h) = 6x.
h!0
h!0
h
(11) Let f be a di↵erentiable function defined on an open interval I. Assume that f 0 = 0 on
I. Prove that f is constant. (5 points)
Proof. Given x, y 2 I such that x < y, f is di↵erentiable on [x, y] since [x, y] ✓ I. Thus
by the Mean Value Theorem, there is c 2 (x, y) such that
f (y)
y
f (x)
= f 0 (c) = 0.
x
Thus f (y) f (x) = 0, so f (x) = f (y). Since f takes the same value on every two
elements of I, f is constant.
Math 3210, Fall 2013
Instructor: Thomas Goller
30 October 2013
(12) Let f (x) = x6 2x 1. Use the Intermediate Value Theorem to prove that f has a real
root, namely that there is c 2 R such that f (c) = 0. (5 points)
Proof. Since f is a polynomial, it is continuous on R. Since f ( 1) = 2, f (0) = 1,
and 1 < 0 < 2, by the Intermediate Value Theorem there is c 2 [ 1, 0] such that
f (c) = 0.