figure should do the trick. You take it from there. Problem 128. Prove the Mean Value Theorem.
figure should do the trick. You take it from there. Problem 128. Prove the Mean Value Theorem.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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#128
![The vertical difference from \( f(x) \) to the secant line, indicated by \( \phi(x) \) in the figure should do the trick. You take it from there.
**Problem 128.** Prove the Mean Value Theorem.
The Mean Value Theorem is extraordinarily useful. Almost all of the properties of the derivative that you used in calculus follow more or less easily from it. For example, the following is true.
**Corollary 3.** If \( f'(x) > 0 \) for every \( x \) in the interval \( (a, b) \), then for every \( c, d \in (a, b) \) where \( d > c \) we have
\[
f(d) > f(c).
\]
That is, \( f \) is increasing on \( (a, b) \).
**Proof:** Suppose \( c \) and \( d \) are as described in the corollary. Then by the Mean Value Theorem there is some number, say \( \alpha \in (c, d) \subseteq (a, b) \) such that
\[
f'(\alpha) = \frac{f(d) - f(c)}{d - c}.
\]
Since \( f'(\alpha) > 0 \) and \( d - c > 0 \) we have \( f(d) - f(c) > 0 \), or \( f(d) > f(c) \).
**Problem 129.** Show that if \( f'(x) < 0 \) for every \( x \) in the interval \( (a, b) \) then \( f \) is decreasing on \( (a, b) \).
**Corollary 4.** Suppose \( f \) is differentiable on some interval \( (a, b) \), \( f' \) is continuous on \( (a, b) \), and that \( f'(c) > 0 \) for some \( c \in (a, b) \). Then there is an interval, \( I \subset (a, b) \), containing \( c \) such that for every \( x, y \) in \( I \) where \( x \geq y \), \( f(x) \geq f(y) \).
**Problem 130.** Prove Cor](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2F022b97c3-67d7-412b-b03a-52904c5e3d6f%2Fkfiem93_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The vertical difference from \( f(x) \) to the secant line, indicated by \( \phi(x) \) in the figure should do the trick. You take it from there.
**Problem 128.** Prove the Mean Value Theorem.
The Mean Value Theorem is extraordinarily useful. Almost all of the properties of the derivative that you used in calculus follow more or less easily from it. For example, the following is true.
**Corollary 3.** If \( f'(x) > 0 \) for every \( x \) in the interval \( (a, b) \), then for every \( c, d \in (a, b) \) where \( d > c \) we have
\[
f(d) > f(c).
\]
That is, \( f \) is increasing on \( (a, b) \).
**Proof:** Suppose \( c \) and \( d \) are as described in the corollary. Then by the Mean Value Theorem there is some number, say \( \alpha \in (c, d) \subseteq (a, b) \) such that
\[
f'(\alpha) = \frac{f(d) - f(c)}{d - c}.
\]
Since \( f'(\alpha) > 0 \) and \( d - c > 0 \) we have \( f(d) - f(c) > 0 \), or \( f(d) > f(c) \).
**Problem 129.** Show that if \( f'(x) < 0 \) for every \( x \) in the interval \( (a, b) \) then \( f \) is decreasing on \( (a, b) \).
**Corollary 4.** Suppose \( f \) is differentiable on some interval \( (a, b) \), \( f' \) is continuous on \( (a, b) \), and that \( f'(c) > 0 \) for some \( c \in (a, b) \). Then there is an interval, \( I \subset (a, b) \), containing \( c \) such that for every \( x, y \) in \( I \) where \( x \geq y \), \( f(x) \geq f(y) \).
**Problem 130.** Prove Cor
![**Real Analysis Educational Content**
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**Problem 126.** Show that \( f'(c) \geq 0 \) and conclude that \( f'(c) = 0 \). ♦
**Problem 127.** Show that if \( f(c) \leq f(x) \) for all \( x \) in some interval \( (a, b) \) then \( f'(c) = 0 \) too. ♦
Many of the most important properties of the derivative follow from what is called the Mean Value Theorem (MVT) which we now state.
**Theorem 22. (The Mean Value Theorem)** Suppose \( f' \) exists for every \( x \in (a, b) \), and \( f \) is continuous on \([a, b]\). Then there is a real number \( c \in (a, b) \) such that
\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]
▲
However, it would be difficult to prove the MVT right now. So we will first state and prove Rolle's Theorem, which can be seen as a special case of the MVT. The proof of the MVT will then follow easily.
Michel Rolle first stated the following theorem in 1691. Given this date and the nature of the theorem it would be reasonable to suppose that Rolle was one of the early developers of calculus but this is not so. In fact, Rolle was disdainful of both Newton and Leibniz's versions of calculus, once deriding them as a collection of "ingenious fallacies." It is a bit ironic that his theorem is so fundamental to the modern development of the calculus he ridiculed.
**Theorem 23. (Rolle’s Theorem)** Suppose \( f' \) exists for every \( x \in (a, b) \), \( f \) is continuous on \([a, b]\), and
\[ f(a) = f(b). \]
Then there is a real number \( c \in (a, b) \) such that
\[ f'(c) = 0. \]
▲
**Proof:** Since \( f \) is continuous on \([a, b]\), we see, by the Extreme Value Theorem, that \( f \) has both a maximum and a minimum on \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc074fc7e-4e48-46f1-bfc4-aee6ba4e6262%2F022b97c3-67d7-412b-b03a-52904c5e3d6f%2F8ons8k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Real Analysis Educational Content**
---
**Problem 126.** Show that \( f'(c) \geq 0 \) and conclude that \( f'(c) = 0 \). ♦
**Problem 127.** Show that if \( f(c) \leq f(x) \) for all \( x \) in some interval \( (a, b) \) then \( f'(c) = 0 \) too. ♦
Many of the most important properties of the derivative follow from what is called the Mean Value Theorem (MVT) which we now state.
**Theorem 22. (The Mean Value Theorem)** Suppose \( f' \) exists for every \( x \in (a, b) \), and \( f \) is continuous on \([a, b]\). Then there is a real number \( c \in (a, b) \) such that
\[
f'(c) = \frac{f(b) - f(a)}{b - a}.
\]
▲
However, it would be difficult to prove the MVT right now. So we will first state and prove Rolle's Theorem, which can be seen as a special case of the MVT. The proof of the MVT will then follow easily.
Michel Rolle first stated the following theorem in 1691. Given this date and the nature of the theorem it would be reasonable to suppose that Rolle was one of the early developers of calculus but this is not so. In fact, Rolle was disdainful of both Newton and Leibniz's versions of calculus, once deriding them as a collection of "ingenious fallacies." It is a bit ironic that his theorem is so fundamental to the modern development of the calculus he ridiculed.
**Theorem 23. (Rolle’s Theorem)** Suppose \( f' \) exists for every \( x \in (a, b) \), \( f \) is continuous on \([a, b]\), and
\[ f(a) = f(b). \]
Then there is a real number \( c \in (a, b) \) such that
\[ f'(c) = 0. \]
▲
**Proof:** Since \( f \) is continuous on \([a, b]\), we see, by the Extreme Value Theorem, that \( f \) has both a maximum and a minimum on \
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