Using the Inverse Function Theorem, what is (r-¹)' (9) given that r(x) = not include "(-¹) (9) =" in your answer.) Provide your answer below: =-3x³ - x² = x + 6? Note that r(-1) = 9. (Do -
Using the Inverse Function Theorem, what is (r-¹)' (9) given that r(x) = not include "(-¹) (9) =" in your answer.) Provide your answer below: =-3x³ - x² = x + 6? Note that r(-1) = 9. (Do -
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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![### Inverse Function Theorem Problem
**Problem Statement:**
Using the Inverse Function Theorem, what is \((r^{-1})'(9)\) given that \(r(x) = -3x^3 - x^2 - x + 6\)? Note that \(r(-1) = 9\). (Do not include \((r^{-1})'(9) =\) in your answer.)
**Provide your answer below:**
[Text Box for Answer]
**Explanation:**
To find \((r^{-1})'(9)\) using the Inverse Function Theorem, follow these steps:
1. **Identify the function \(r(x)\)** and verify the condition provided.
- Function: \( r(x) = -3x^3 - x^2 - x + 6 \)
- Given: \( r(-1) = 9 \)
2. **Find the derivative of \(r(x)\), denoted as \( r'(x) \)**.
3. **Evaluate \( r'(x) \) at \( x = -1 \)** because \( r(-1) = 9 \).
4. **Apply the Inverse Function Theorem**, which states:
\[
(r^{-1})'(y) = \frac{1}{r'(x)} \quad \text{where} \quad r(x) = y.
\]
In this case, \((r^{-1})'(9) = \frac{1}{r'(-1)}\).
**Derivative Calculation:**
Calculate \( r'(x) \):
\[ r(x) = -3x^3 - x^2 - x + 6 \]
Applying the power rule:
\[ r'(x) = -9x^2 - 2x - 1 \]
**Evaluate at \( x = -1 \):**
\[ r'(-1) = -9(-1)^2 - 2(-1) - 1 \]
\[ r'(-1) = -9 + 2 - 1 \]
\[ r'(-1) = -8 \]
**Inverse Function Theorem Application:**
\[ (r^{-1})'(9) = \frac{1}{r'(-1)} = \frac{1}{-8} = -\frac{1}{8} \]
Therefore, the answer is:
\[ \boxed](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb47bac88-cba5-40e5-9f3f-db7316eb95a3%2F60b54b8f-127e-44e3-a936-f935452fe437%2Fms2rz18_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Inverse Function Theorem Problem
**Problem Statement:**
Using the Inverse Function Theorem, what is \((r^{-1})'(9)\) given that \(r(x) = -3x^3 - x^2 - x + 6\)? Note that \(r(-1) = 9\). (Do not include \((r^{-1})'(9) =\) in your answer.)
**Provide your answer below:**
[Text Box for Answer]
**Explanation:**
To find \((r^{-1})'(9)\) using the Inverse Function Theorem, follow these steps:
1. **Identify the function \(r(x)\)** and verify the condition provided.
- Function: \( r(x) = -3x^3 - x^2 - x + 6 \)
- Given: \( r(-1) = 9 \)
2. **Find the derivative of \(r(x)\), denoted as \( r'(x) \)**.
3. **Evaluate \( r'(x) \) at \( x = -1 \)** because \( r(-1) = 9 \).
4. **Apply the Inverse Function Theorem**, which states:
\[
(r^{-1})'(y) = \frac{1}{r'(x)} \quad \text{where} \quad r(x) = y.
\]
In this case, \((r^{-1})'(9) = \frac{1}{r'(-1)}\).
**Derivative Calculation:**
Calculate \( r'(x) \):
\[ r(x) = -3x^3 - x^2 - x + 6 \]
Applying the power rule:
\[ r'(x) = -9x^2 - 2x - 1 \]
**Evaluate at \( x = -1 \):**
\[ r'(-1) = -9(-1)^2 - 2(-1) - 1 \]
\[ r'(-1) = -9 + 2 - 1 \]
\[ r'(-1) = -8 \]
**Inverse Function Theorem Application:**
\[ (r^{-1})'(9) = \frac{1}{r'(-1)} = \frac{1}{-8} = -\frac{1}{8} \]
Therefore, the answer is:
\[ \boxed
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