### 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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Description: ### HW 8.5 #1:**Problem Statement:**Suppose $ f $ and $ g $ are continuous on the interval $[a, b]$ and that $ f(a) g(b) $. Prove that there is a real number $ x_0 \in (a, b) $ such that $ f(x_0) = g(x_0) $.**Hint

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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

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