Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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, \((f^{-1})'(1) = \frac{1}{6}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c444b80-0f04-4660-83de-5c28618d5488%2Fef2ad14e-ada4-44bf-8183-4f05fd4d35e9%2Fmy9vqd_processed.jpeg&w=3840&q=75)
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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