MAT 3749 2.2 Part II Handout
The Intermediate Value Theorem (Statement 1)
Suppose that f is continuous on  a, b and let N be any number between f (a ) and f (b ) ,
where f (a )  f (b) . Then there exists a number c in  a, b  such that f (c)  N .
Recall: Continuity - The e-d Definition
A function f defined on an open interval contains c is continuous at c if
  0 ,   0 such that f  x   f  c    whenever x  c   .
Continuous from the Right
A function f defined on an open interval c, b  is continuous from the right at c if
  0 ,   0 such that f  x   f  c    whenever x  c, c    .
Recall: Least Upper Bounds
A real number M is the least upper bound, or Supremum, of a set E 
1. M is an upper bound of E;
2. If N  M , then N is not an upper bound of E.
Recall: The Completeness Axiom
Let E be a non-empty subset of .
1. If E is bounded above, then E has a least upper bound.
2. If E is bounded below, then E has a greatest lower bound.
Last Edit: 3/9/2016 3:39 AM
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The Intermediate Value Theorem (Statement 2)
Suppose that f is continuous on  a, b and that f (a ) , f (b ) are with different signs. Then
there exists a number c in  a, b  such that f (c)  0 .
f  x
f b
a
c
0
x
b
f a
Analysis
WLOG (Without Loss of Generality), assume f  a   0  f  b  .
The first question is how to choose c . Let us look at two different scenarios.
f  x
f  x
f b
f b
a
0
f a
S
c
b
x
a
S
0
cb
x
f a
A possible way to choose c is such that c is the last point where f  x   0 .
Let S   x   a, b  , f  x   0 . We want to choose c such that c  sup S .
sup S
exists?
c   a, b ?
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To prove that f  c   0 , we prove instead f (c)  0 and f (c)  0 .
Suppose f (c)  0 .
We need a contradiction.
Suppose f (c)  0 .
We need a contradiction.
f  x
f  x
c x1
x
x2 c
x
c   a, b  ?
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Lemma 1
Let f be continuous on  a, b and c   a, b  such that f (c)  0 .
Then   0 such that f ( x)  0 x   c, c    .
Analysis
Proof
f  x
c
c 
x
Lemma 2
Let f be continuous on  a, b and c   a, b such that f (c)  0 .
Then   0 such that f ( x)  0 x   c   , c  .
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The Intermediate Value Theorem (Statement 2)
Suppose that f is continuous on  a, b and that f (a ) , f (b ) are with different signs.
Then there exists a number c in  a, b  such that f (c)  0 .
Proof
WLOG (without Loss of Generality), assume f  a   0  f  b  .
Let S   x   a, b  , f  x   0 .
1. f  a   0  a  S  S  
2. x  S , x  b  S is bounded above by b .
So, sup S exists. Let c be such that c  sup S .
1. a  S and c is an upper bound of S  a  c
2. b is an upper bound of S and c  sup S  c  b
Therefore, c   a, b .
Suppose f (c)  0 . In particular, c  b .
Now, f is continuous on  a, b and c   a, b  .
By lemma 1, Then 1  0 such that f ( x)  0 x   c, c  1  .
Choose any x1   c, c  1  .
Then x1  S and c  x1 which is a contradiction that c is an upper bound of S .
Thus, f (c)  0 .
Suppose f (c)  0 . In particular, c  a and c  S .
Now, f is continuous on  a, b and c   a, b .
By lemma 2,  2  0 such that f ( x)  0 x   c   2 , c  .
Choose any x2   c   2 , c  .
Since x2  c which is the least upper bound of S , x2 is not an upper bound of S .
Thus z  S such that x2  z . Also, z  c since c is an upper bound of S .
So, z   c   2 , c   f  z   0 which is a contradiction that z  S .
Thus, f (c)  0 .
As a result, we prove that c   a, b such that f (c)  0 .
Since f  a   0  f  b  , c  a or b . Thus, c   a, b  .
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