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
![**Topic: Calculus - Finding the Equation of a Tangent Line**
**Problem:**
Find the equation of the tangent line to the graph of \( f(x) = -2(3x - 1)^3 \) at \( x = 1 \). Show work or explain your reasoning.
**Solution:**
To find the equation of the tangent line, we need:
1. **The point** on the curve at \( x = 1 \).
2. **The slope** of the tangent line at that point.
**Step 1: Find the Point on the Curve**
Substitute \( x = 1 \) into the function:
\[
f(x) = -2(3x - 1)^3
\]
\[
f(1) = -2(3(1) - 1)^3 = -2(3 - 1)^3 = -2 \times 2^3 = -2 \times 8 = -16
\]
Thus, the point on the curve is \( (1, -16) \).
**Step 2: Find the Slope of the Tangent Line**
The slope of the tangent line is given by the derivative of \( f(x) \). First, find \( f'(x) \):
Using the chain rule:
\[
f(x) = -2(3x - 1)^3
\]
\[
f'(x) = -2 \cdot 3 \cdot 3(3x - 1)^2 = -18(3x - 1)^2
\]
Now, find \( f'(1) \):
\[
f'(1) = -18(3(1) - 1)^2 = -18(2)^2 = -18 \times 4 = -72
\]
**Step 3: Write the Equation of the Tangent Line**
Using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
With point \( (1, -16) \) and slope \( -72 \):
\[
y + 16 = -72(x - 1)
\]
Simplifying:
\[
y + 16 = -72x + 72
\]
\[
y = -72x + 72 - 16
\]
\[
y = -72x + 56
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F236de4c3-f237-402b-b0e0-efa1a8147b53%2F16ddff82-76ec-4e3c-8839-671b0a75d38b%2Fe47llq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Calculus - Finding the Equation of a Tangent Line**
**Problem:**
Find the equation of the tangent line to the graph of \( f(x) = -2(3x - 1)^3 \) at \( x = 1 \). Show work or explain your reasoning.
**Solution:**
To find the equation of the tangent line, we need:
1. **The point** on the curve at \( x = 1 \).
2. **The slope** of the tangent line at that point.
**Step 1: Find the Point on the Curve**
Substitute \( x = 1 \) into the function:
\[
f(x) = -2(3x - 1)^3
\]
\[
f(1) = -2(3(1) - 1)^3 = -2(3 - 1)^3 = -2 \times 2^3 = -2 \times 8 = -16
\]
Thus, the point on the curve is \( (1, -16) \).
**Step 2: Find the Slope of the Tangent Line**
The slope of the tangent line is given by the derivative of \( f(x) \). First, find \( f'(x) \):
Using the chain rule:
\[
f(x) = -2(3x - 1)^3
\]
\[
f'(x) = -2 \cdot 3 \cdot 3(3x - 1)^2 = -18(3x - 1)^2
\]
Now, find \( f'(1) \):
\[
f'(1) = -18(3(1) - 1)^2 = -18(2)^2 = -18 \times 4 = -72
\]
**Step 3: Write the Equation of the Tangent Line**
Using the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]
With point \( (1, -16) \) and slope \( -72 \):
\[
y + 16 = -72(x - 1)
\]
Simplifying:
\[
y + 16 = -72x + 72
\]
\[
y = -72x + 72 - 16
\]
\[
y = -72x + 56
\
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