Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
**Problem Statement:** Given that \[ f(x) = x^{11} h(x) \] and \[ h(-1) = 2 \] \[ h'(-1) = 5 \] Calculate \( f'(-1) \). **Hint:** Use the product rule and the power rule. --- **Solution Reference:** To solve for \( f'(-1) \), apply the product rule for differentiation: if \( f(x) = u(x)v(x) \), then \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] For this problem: - Let \( u(x) = x^{11} \) and \( v(x) = h(x) \). - Therefore, \( u'(x) = 11x^{10} \) using the power rule. Substitute into the product rule formula: \[ f'(x) = 11x^{10}h(x) + x^{11}h'(x) \] Evaluate \( f'(-1) \): \[ f'(-1) = 11(-1)^{10}h(-1) + (-1)^{11}h'(-1) \] Since: \[ (-1)^{10} = 1 \] and \[ (-1)^{11} = -1 \] Substitute the given values: \[ f'(-1) = 11(1)(2) + (-1)(5) \] \[ f'(-1) = 22 - 5 \] \[ f'(-1) = 17 \] So, \( f'(-1) = 17 \).
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Transcribed Image Text:**Problem Statement:** Given that \[ f(x) = x^{11} h(x) \] and \[ h(-1) = 2 \] \[ h'(-1) = 5 \] Calculate \( f'(-1) \). **Hint:** Use the product rule and the power rule. --- **Solution Reference:** To solve for \( f'(-1) \), apply the product rule for differentiation: if \( f(x) = u(x)v(x) \), then \[ f'(x) = u'(x)v(x) + u(x)v'(x) \] For this problem: - Let \( u(x) = x^{11} \) and \( v(x) = h(x) \). - Therefore, \( u'(x) = 11x^{10} \) using the power rule. Substitute into the product rule formula: \[ f'(x) = 11x^{10}h(x) + x^{11}h'(x) \] Evaluate \( f'(-1) \): \[ f'(-1) = 11(-1)^{10}h(-1) + (-1)^{11}h'(-1) \] Since: \[ (-1)^{10} = 1 \] and \[ (-1)^{11} = -1 \] Substitute the given values: \[ f'(-1) = 11(1)(2) + (-1)(5) \] \[ f'(-1) = 22 - 5 \] \[ f'(-1) = 17 \] So, \( f'(-1) = 17 \).
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