Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
Find \((f^{-1})'(a)\). Given: \[ f(x) = 3x^3 + 4x^2 + 6x + 1 \] \[ a = 1 \] Determine \((f^{-1})'(a)\). **Solution:** To find \((f^{-1})'(a)\), we can use the formula for the derivative of an inverse function: \[ (f^{-1})'(a) = \frac{1}{f'(b)} \] where \(b\) is such that \(f(b) = a\). 1. **Compute \(f'(x)\):** \[ f'(x) = \frac{d}{dx}(3x^3 + 4x^2 + 6x + 1) = 9x^2 + 8x + 6 \] 2. **Solve \(f(x) = 1\):** Set the equation \(3x^3 + 4x^2 + 6x + 1 = 1\). \[ 3x^3 + 4x^2 + 6x = 0 \] Factor the equation: \[ x(3x^2 + 4x + 6) = 0 \] Solutions: \[ x = 0 \quad (\text{Other factors may need further analysis to verify additional roots}) \] 3. **Substitute \(x = 0\) in \(f'(x)\):** \[ f'(0) = 9(0)^2 + 8(0) + 6 = 6 \] 4. **Compute \((f^{-1})'(1)\):** \[ (f^{-1})'(1) = \frac{1}{6} \] Hence, \((f^{-1})'(1) = \frac{1}{6}\).
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Transcribed Image Text:Find \((f^{-1})'(a)\). Given: \[ f(x) = 3x^3 + 4x^2 + 6x + 1 \] \[ a = 1 \] Determine \((f^{-1})'(a)\). **Solution:** To find \((f^{-1})'(a)\), we can use the formula for the derivative of an inverse function: \[ (f^{-1})'(a) = \frac{1}{f'(b)} \] where \(b\) is such that \(f(b) = a\). 1. **Compute \(f'(x)\):** \[ f'(x) = \frac{d}{dx}(3x^3 + 4x^2 + 6x + 1) = 9x^2 + 8x + 6 \] 2. **Solve \(f(x) = 1\):** Set the equation \(3x^3 + 4x^2 + 6x + 1 = 1\). \[ 3x^3 + 4x^2 + 6x = 0 \] Factor the equation: \[ x(3x^2 + 4x + 6) = 0 \] Solutions: \[ x = 0 \quad (\text{Other factors may need further analysis to verify additional roots}) \] 3. **Substitute \(x = 0\) in \(f'(x)\):** \[ f'(0) = 9(0)^2 + 8(0) + 6 = 6 \] 4. **Compute \((f^{-1})'(1)\):** \[ (f^{-1})'(1) = \frac{1}{6} \] Hence, \((f^{-1})'(1) = \frac{1}{6}\).
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