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...
Related questions
Question
![**Problem Statement:**
Find the equation of the tangent line to the graph of \( f(x) = 4x^3 + 7x^2 - 8x - 6 \) at \( x = -1 \).
---
**Solution Approach:**
1. **Find the derivative \( f'(x) \)**
The derivative \( f'(x) \) represents the slope of the tangent line to the curve at any point \( x \).
2. **Evaluate \( f'(x) \) at \( x = -1 \)**
Determine the slope of the tangent line at the given point by substituting \( x = -1 \) into \( f'(x) \).
3. **Find \( f(-1) \)**
Substitute \( x = -1 \) into the original function \( f(x) \) to find the \( y \)-coordinate of the point of tangency.
4. **Use the point-slope form of a line**
Use the slope found in Step 2 and the point \((-1, f(-1))\) from Step 3 to write the equation of the tangent line in the form:
\[
y - f(-1) = f'(-1)(x + 1)
\]
---
**Final Answer:**
The equation of the tangent line is displayed in the box below:
\[ y = \text{[Equation Here]} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0cd85eb8-5592-4bd0-9b53-cd63c8c5d902%2Feeb704a2-eb86-447f-93ba-7f98f5dfa771%2Fo4spw7_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of the tangent line to the graph of \( f(x) = 4x^3 + 7x^2 - 8x - 6 \) at \( x = -1 \).
---
**Solution Approach:**
1. **Find the derivative \( f'(x) \)**
The derivative \( f'(x) \) represents the slope of the tangent line to the curve at any point \( x \).
2. **Evaluate \( f'(x) \) at \( x = -1 \)**
Determine the slope of the tangent line at the given point by substituting \( x = -1 \) into \( f'(x) \).
3. **Find \( f(-1) \)**
Substitute \( x = -1 \) into the original function \( f(x) \) to find the \( y \)-coordinate of the point of tangency.
4. **Use the point-slope form of a line**
Use the slope found in Step 2 and the point \((-1, f(-1))\) from Step 3 to write the equation of the tangent line in the form:
\[
y - f(-1) = f'(-1)(x + 1)
\]
---
**Final Answer:**
The equation of the tangent line is displayed in the box below:
\[ y = \text{[Equation Here]} \]
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