The Extreme Value Theorem (EVT)
EVT: If f is continuous on a closed interval [a, b], then there exists values
M and m in the interval [a, b] such that f (M) is the maximum value that
f (x) attains on [a, b], and f (m) is the minimum value that f (x) attains on
[a, b].
Math 101-Calculus 1 (Sklensky)
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The Extreme Value Theorem (EVT)
I
Example 1: Suppose f (x) is given by the function shown below.
20
2
4
6
8
10
-20
-40
-60
I
I
I
f is continuous on the interval [0, 11]
The EVT ⇒ At some point between x = 0 and x = 11, the graph
reaches a maximum height.
The EVT ⇒ At some point between x = 0 and x = 11, the graph
reaches a minimum height.
Math 101-Calculus 1 (Sklensky)
In-Class Work
February 9, 2015
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The Extreme Value Theorem (EVT)
I
Example 1: Suppose f (x) is given by the function shown below.
20
2
4
6
M
8
m 10
-20
-40
-60
I
I
I
f is continuous on the interval [0, 11]
The EVT ⇒ At some point between x = 0 and x = 11, the graph
reaches a maximum height.
The EVT ⇒ At some point between x = 0 and x = 11, the graph
reaches a minimum height.
Math 101-Calculus 1 (Sklensky)
In-Class Work
February 9, 2015
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The Intermediate Value Theorem (IVT)
IVT: If f is continuous on a closed interval [a, b], then for any number K
strictly between f (a) and f (b), there exists at least one number c ∈ [a, b]
for which f (c) = K .
I
Hypotheses: (that is, conditions that must be satisfied in order to
apply the theorem)
I
I
f is continuous on [a, b]
Conclusion: (that is, what we can conclude if the hypotheses are
satisfied)
I
For every number W between f (a) and f (b), there is some number c
between a and b so that f (c) = W .
Math 101-Calculus 1 (Sklensky)
In-Class Work
February 9, 2015
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Illustrating the IVT:
f(a)
f(x)
f(b)
b
a
Math 101-Calculus 1 (Sklensky)
In-Class Work
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Illustrating the IVT:
f(a)
W lies
between
f(a) and
f(b)
W
f(x)
f(b)
a
c
b
There is
a c such
that f(c)=W
Math 101-Calculus 1 (Sklensky)
In-Class Work
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Illustrating the IVT:
W1
f(a)
If W doesn!t lie
between f(a)
and f(b), there
may or may
not be such a
c.
f(x)
f(b)
W2
a
Math 101-Calculus 1 (Sklensky)
c
b
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In Class Work
1. Let f (x) = 14 sin(3x) + 2x 2 − 4x 3 . Use the IVT to show that f (x)
has a root between x = −2 and x = 2.
1
. Use the IVT to show that f (x) has a root between
x −2
x = 1 and x = 3.
2. (a) Let f (x) =
(b) Find the exact value of the root by solving f (x) = 0. What goes
wrong?
(c) Reconcile your answers to parts (a) and (b).
Math 101-Calculus 1 (Sklensky)
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Solutions:
1. Let f (x) = 14 sin(3x) + 2x 2 − 4x 3 . Use the IVT to show that f (x)
has a root between x = −2 and x = 2.
f (−2) = 14 sin(−6) + 8 − 4(−8) = 14 sin(−6) + 8 + 32 > 0
f (2) = 14 sin(6) + 8 − 32 < 0
Because f is continuous on [−2, 2] and because 0 is between f (−2)
and f (2), there must be some c ∈ [−2, 2] such that f (c) = 0.
Therefore f has a root between x = −2 and x = 2.
Math 101-Calculus 1 (Sklensky)
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Solutions:
2. Let f (x) =
1
.
x −2
(a) Use the IVT to show that f (x) has a root between x = 1 and x = 3.
f (1) = −1
f (3) = 1
Since f (1) < 0 and f (3) > 0, it seems that f has a root between x = 1
and x = 3.
(b) Find the exact value of the root by solving f (x) = 0. What goes
wrong?
1
= 0 =⇒ (x − 2)
x −2
1
x −2
= (x − 2) 0 =⇒ 1 = 0!!
Nonsensical conclusion =⇒ Original set-up must have made no sense
=⇒ there is no root!
Math 101-Calculus 1 (Sklensky)
In-Class Work
February 9, 2015
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2. (continued)
(c) Reconcile your answers to parts (a) and (b).
How can there not be a root – we used the IVT to show a root must
exist!!
But did we? Did we ever check to see whether the hypotheses of the
theorem apply?
Is f (x) continuous on [1, 3]?
No – f (x) =
Math 101-Calculus 1 (Sklensky)
1
is not defined at x = 2.
x −2
In-Class Work
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