Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question

3. Let f(x) = 1/x2
. Then f(−1) = f(1). Explain why it is not possible to use Rolle’s Theorem to
show that f
0
(c) = 0 for some c in (−1, 1).

**3. Let \( f(x) = \frac{1}{x^2} \). Then \( f(-1) = f(1) \). Explain why it is *not* possible to use Rolle’s Theorem to show that \( f'(c) = 0 \) for some \( c \) in \((-1, 1)\).** ### Explanation: Rolle’s Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in the interval \((a, b)\) such that \( f'(c) = 0 \). - **Continuity Requirement:** The function \( f(x) = \frac{1}{x^2} \) is not continuous on the interval \([-1, 1]\) because it is not defined at \( x = 0 \). The function has a discontinuity or an asymptote at \( x = 0 \). - **Differentiability Requirement:** Since the function is not continuous on \([-1, 1]\), it cannot be differentiable on \((-1, 1)\). Thus, due to the discontinuity at \( x = 0 \), the function fails to meet the continuity (and, consequently, the differentiability) requirements necessary for applying Rolle's Theorem.
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Transcribed Image Text:**3. Let \( f(x) = \frac{1}{x^2} \). Then \( f(-1) = f(1) \). Explain why it is *not* possible to use Rolle’s Theorem to show that \( f'(c) = 0 \) for some \( c \) in \((-1, 1)\).** ### Explanation: Rolle’s Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in the interval \((a, b)\) such that \( f'(c) = 0 \). - **Continuity Requirement:** The function \( f(x) = \frac{1}{x^2} \) is not continuous on the interval \([-1, 1]\) because it is not defined at \( x = 0 \). The function has a discontinuity or an asymptote at \( x = 0 \). - **Differentiability Requirement:** Since the function is not continuous on \([-1, 1]\), it cannot be differentiable on \((-1, 1)\). Thus, due to the discontinuity at \( x = 0 \), the function fails to meet the continuity (and, consequently, the differentiability) requirements necessary for applying Rolle's Theorem.
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