I have to explain each step of the solved problems in the picture.
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter7: Distance And Approximation
Section7.3: Least Squares Approximation
Problem 32EQ
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I have to explain each step of the solved problems in the picture.
![Mean value theorems
4.19.
Prove Rolle's theorem.
Case 1: f(x) =0 in [a, b]. Thenf'(x) = 0 for all x in (a, b).
Case 2: f(x) # 0 in [a, b]. Since f(x) is continuous, there are points at which f(x) attains its maximum and
minimum values, denoted by M and m, respectively (see Problem 3.34).
Since f(x) # 0, at least one of the values M, m is not zero. Suppose, for example, M # 0 and that f() =
M (see Figure 4.9). For this case, f(Ě + h) < f().
f(x)
M
Figure 4.9
f(Ë + h) – f(§)
If h >0, then-
O and
h
f(Ë +h) – f(E)
lim
(1)
h0+
h
f(Ë + h)- ƒ(E)
If h <0, then
20 and
h
fË + h) – f(E).
lim
(2)
h→0-
h
But, by hypothesis, f(x) has a derivative at all points in (a, b). Then the right-hand derivative (1) must
be equal to the left-hand derivative (2). This can happen only if they are both equal to zero, in which case
f'(E) = 0 as required.
A similar argument can be used in case M = 0 and m + 0.
3.34.
Prove Theorem 10, Page 52.
Given any e > 0, we can find x such that M - f(x) <e by definition of the I.u.b. M.
1
1
so that
1
Then
>
is not bounded and, hence, cannot be continuous in view of
M – f(x) E
M - f(x)
Theorem 4, Page 52. However, if we suppose that f(x) # M, then, since M - f(x) is continuous, by hypothesis
1
we must havę
also continuous. In view of this contradiction, we must have f(x) = M for at least
M- f(x)
one value of x in the interval.
Similarly, we can show that there exists an x in the interval such that f(x) = m (Problem 3.93).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2Fa91e74c8-ad82-4fd0-82b3-b0cc4d98bbc0%2Fhm5acht_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Mean value theorems
4.19.
Prove Rolle's theorem.
Case 1: f(x) =0 in [a, b]. Thenf'(x) = 0 for all x in (a, b).
Case 2: f(x) # 0 in [a, b]. Since f(x) is continuous, there are points at which f(x) attains its maximum and
minimum values, denoted by M and m, respectively (see Problem 3.34).
Since f(x) # 0, at least one of the values M, m is not zero. Suppose, for example, M # 0 and that f() =
M (see Figure 4.9). For this case, f(Ě + h) < f().
f(x)
M
Figure 4.9
f(Ë + h) – f(§)
If h >0, then-
O and
h
f(Ë +h) – f(E)
lim
(1)
h0+
h
f(Ë + h)- ƒ(E)
If h <0, then
20 and
h
fË + h) – f(E).
lim
(2)
h→0-
h
But, by hypothesis, f(x) has a derivative at all points in (a, b). Then the right-hand derivative (1) must
be equal to the left-hand derivative (2). This can happen only if they are both equal to zero, in which case
f'(E) = 0 as required.
A similar argument can be used in case M = 0 and m + 0.
3.34.
Prove Theorem 10, Page 52.
Given any e > 0, we can find x such that M - f(x) <e by definition of the I.u.b. M.
1
1
so that
1
Then
>
is not bounded and, hence, cannot be continuous in view of
M – f(x) E
M - f(x)
Theorem 4, Page 52. However, if we suppose that f(x) # M, then, since M - f(x) is continuous, by hypothesis
1
we must havę
also continuous. In view of this contradiction, we must have f(x) = M for at least
M- f(x)
one value of x in the interval.
Similarly, we can show that there exists an x in the interval such that f(x) = m (Problem 3.93).
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