HW 8.5 #1: Suppose f and g are continuous on the interval [a, b] and that f(a) < g(a) and f(b) > g(b). Prove that there is a real number ro E (a, b) such that (Ox)6 = (0x)S S(r) – g(x) and apply the Intermediate Value (HINT: Define a function h(r) Theorem.)
HW 8.5 #1: Suppose f and g are continuous on the interval [a, b] and that f(a) < g(a) and f(b) > g(b). Prove that there is a real number ro E (a, b) such that (Ox)6 = (0x)S S(r) – g(x) and apply the Intermediate Value (HINT: Define a function h(r) Theorem.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### HW 8.5 #1:
**Problem Statement:**
Suppose \( f \) and \( g \) are continuous on the interval \([a, b]\) and that \( f(a) < g(a) \) and \( f(b) > g(b) \). Prove that there is a real number \( x_0 \in (a, b) \) such that \( f(x_0) = g(x_0) \).
**Hint:**
Define a function \( h(x) = f(x) - g(x) \) and apply the Intermediate Value Theorem.
### Detailed Explanation:
The problem requires proving that two continuous functions, \( f \) and \( g \), intersect at some point within the interval \((a, b)\). This can be approached using the concept of the Intermediate Value Theorem (IVT) which is a fundamental theorem in calculus.
**Steps:**
1. **Define a New Function:**
- Define \( h(x) = f(x) - g(x) \).
- Since \( f \) and \( g \) are continuous, and the difference of continuous functions is continuous, \( h \) is continuous on \([a, b]\).
2. **Evaluate at the Endpoints:**
- \( h(a) = f(a) - g(a) \). Given \( f(a) < g(a) \), we have \( h(a) < 0 \).
- \( h(b) = f(b) - g(b) \). Given \( f(b) > g(b) \), we have \( h(b) > 0 \).
3. **Apply the Intermediate Value Theorem:**
- \( h(x) \) is continuous, and since it changes signs from \( a \) to \( b \) (i.e., \( h(a) < 0 \) to \( h(b) > 0 \)), by the IVT, there exists some \( x_0 \in (a, b) \) where \( h(x_0) = 0 \).
4. **Conclusion:**
- Thus, \( h(x_0) = 0 \implies f(x_0) - g(x_0) = 0 \implies f(x_0) = g(x_0) \).
This proves there is a real number \( x_0 \) in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87a5eb21-df18-43d0-b53f-a372e6adca02%2Fb7ac29a9-99e7-454a-8547-c2f790b2afa0%2F2cvhri_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### HW 8.5 #1:
**Problem Statement:**
Suppose \( f \) and \( g \) are continuous on the interval \([a, b]\) and that \( f(a) < g(a) \) and \( f(b) > g(b) \). Prove that there is a real number \( x_0 \in (a, b) \) such that \( f(x_0) = g(x_0) \).
**Hint:**
Define a function \( h(x) = f(x) - g(x) \) and apply the Intermediate Value Theorem.
### Detailed Explanation:
The problem requires proving that two continuous functions, \( f \) and \( g \), intersect at some point within the interval \((a, b)\). This can be approached using the concept of the Intermediate Value Theorem (IVT) which is a fundamental theorem in calculus.
**Steps:**
1. **Define a New Function:**
- Define \( h(x) = f(x) - g(x) \).
- Since \( f \) and \( g \) are continuous, and the difference of continuous functions is continuous, \( h \) is continuous on \([a, b]\).
2. **Evaluate at the Endpoints:**
- \( h(a) = f(a) - g(a) \). Given \( f(a) < g(a) \), we have \( h(a) < 0 \).
- \( h(b) = f(b) - g(b) \). Given \( f(b) > g(b) \), we have \( h(b) > 0 \).
3. **Apply the Intermediate Value Theorem:**
- \( h(x) \) is continuous, and since it changes signs from \( a \) to \( b \) (i.e., \( h(a) < 0 \) to \( h(b) > 0 \)), by the IVT, there exists some \( x_0 \in (a, b) \) where \( h(x_0) = 0 \).
4. **Conclusion:**
- Thus, \( h(x_0) = 0 \implies f(x_0) - g(x_0) = 0 \implies f(x_0) = g(x_0) \).
This proves there is a real number \( x_0 \) in
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