Let f be a real-valued continuous and differentiable function. Let function g be defined by g(x) = f(|x| + 2). A student presents the following proof to show that there exists a real number c € (-1, 1) such that g/(c) = 0. (1) Since f is a continuous function, so is g over the interval [-1, 1]. (II) Since f is differentiable, so is g over the interval (-1, 1). (III) It is evident from the definition of g that g(-1) = g(1). (IV) If the above conditions hold, then by Rolle's theorem, there exists dg(x) c = (-1, 1) such that gl (c) dx |x=c Which statement about this proof is correct? = 0. Step (1) does not hold, and hence Rolle's theorem does not apply. Step (II) does not hold, and hence Rolle's theorem does not apply. Step (III) does not hold, and hence Rolle's theorem does not apply. Step (IV) does not hold, and hence the conclusion is false. The proof is completely correct, and the conclusion holds.
Let f be a real-valued continuous and differentiable function. Let function g be defined by g(x) = f(|x| + 2). A student presents the following proof to show that there exists a real number c € (-1, 1) such that g/(c) = 0. (1) Since f is a continuous function, so is g over the interval [-1, 1]. (II) Since f is differentiable, so is g over the interval (-1, 1). (III) It is evident from the definition of g that g(-1) = g(1). (IV) If the above conditions hold, then by Rolle's theorem, there exists dg(x) c = (-1, 1) such that gl (c) dx |x=c Which statement about this proof is correct? = 0. Step (1) does not hold, and hence Rolle's theorem does not apply. Step (II) does not hold, and hence Rolle's theorem does not apply. Step (III) does not hold, and hence Rolle's theorem does not apply. Step (IV) does not hold, and hence the conclusion is false. The proof is completely correct, and the conclusion holds.
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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![**Problem Statement:**
Let \( f \) be a real-valued continuous and differentiable function. Let function \( g \) be defined by \( g(x) = f(|x| + 2) \). A student presents the following proof to show that there exists a real number \( c \in (-1, 1) \) such that \( g'(c) = 0 \).
1. Since \( f \) is a continuous function, so is \( g \) over the interval \( [-1, 1] \).
2. Since \( f \) is differentiable, so is \( g \) over the interval \( (-1, 1) \).
3. It is evident from the definition of \( g \) that \( g(-1) = g(1) \).
4. If the above conditions hold, then by Rolle's theorem, there exists \( c \in (-1, 1) \) such that
\[
g'(c) = \left. \frac{dg(x)}{dx} \right|_{x=c} = 0.
\]
**Question:**
Which statement about this proof is correct?
- Step (I) does not hold, and hence Rolle's theorem does not apply.
- Step (II) does not hold, and hence Rolle's theorem does not apply.
- Step (III) does not hold, and hence Rolle's theorem does not apply.
- Step (IV) does not hold, and hence the conclusion is false.
- The proof is completely correct, and the conclusion holds.
**Analysis of Proof:**
1. **Step (I)**: Considers whether \( g(x) = f(|x| + 2) \) is continuous over the interval \([-1, 1]\). Given that \( f \) is continuous and the absolute value is continuous, the composition \( g(x) \) should also be continuous over \([-1, 1]\). Hence, Step (I) holds.
2. **Step (II)**: Considers whether \( g(x) \) is differentiable over the interval \((-1, 1)\). Function \( f \) is differentiable, but since \( g(x) = f(|x| + 2) \), we need to check the differentiability of this composition. The absolute value function \( |x| \) is not differentiable at](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49dd86f5-e3e4-495a-9b19-f3d039df6a7c%2Fd73812c2-8b2c-4435-872d-8aae98baecc4%2Fbjjzjvk_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( f \) be a real-valued continuous and differentiable function. Let function \( g \) be defined by \( g(x) = f(|x| + 2) \). A student presents the following proof to show that there exists a real number \( c \in (-1, 1) \) such that \( g'(c) = 0 \).
1. Since \( f \) is a continuous function, so is \( g \) over the interval \( [-1, 1] \).
2. Since \( f \) is differentiable, so is \( g \) over the interval \( (-1, 1) \).
3. It is evident from the definition of \( g \) that \( g(-1) = g(1) \).
4. If the above conditions hold, then by Rolle's theorem, there exists \( c \in (-1, 1) \) such that
\[
g'(c) = \left. \frac{dg(x)}{dx} \right|_{x=c} = 0.
\]
**Question:**
Which statement about this proof is correct?
- Step (I) does not hold, and hence Rolle's theorem does not apply.
- Step (II) does not hold, and hence Rolle's theorem does not apply.
- Step (III) does not hold, and hence Rolle's theorem does not apply.
- Step (IV) does not hold, and hence the conclusion is false.
- The proof is completely correct, and the conclusion holds.
**Analysis of Proof:**
1. **Step (I)**: Considers whether \( g(x) = f(|x| + 2) \) is continuous over the interval \([-1, 1]\). Given that \( f \) is continuous and the absolute value is continuous, the composition \( g(x) \) should also be continuous over \([-1, 1]\). Hence, Step (I) holds.
2. **Step (II)**: Considers whether \( g(x) \) is differentiable over the interval \((-1, 1)\). Function \( f \) is differentiable, but since \( g(x) = f(|x| + 2) \), we need to check the differentiability of this composition. The absolute value function \( |x| \) is not differentiable at
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