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
DERIVATIVES
![**12. Find the equation of the tangent line to \( f(x) = x^3 \) at the point \( P(1, 1) \).**
To find the equation of the tangent line at a given point on a curve, we need to determine the slope of the tangent line and use the point-slope form of a line equation.
**Steps to solve:**
1. **Find the Derivative of the Function:**
The derivative \( f'(x) \) of \( f(x) = x^3 \) gives the slope of the tangent to the curve at any point \( x \).
\[
f'(x) = 3x^2
\]
2. **Calculate the Slope at the Specific Point:**
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \( P(1, 1) \).
\[
f'(1) = 3(1)^2 = 3
\]
3. **Use the Point-Slope Form of a Line:**
The point-slope form of the equation of a line is:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is a point on the line. Using \( m = 3 \) and \( (x_1, y_1) = (1, 1) \):
\[
y - 1 = 3(x - 1)
\]
4. **Simplify the Equation:**
Expand and rearrange the equation to standard form:
\[
y - 1 = 3x - 3
\]
\[
y = 3x - 2
\]
Therefore, the equation of the tangent line to \( f(x) = x^3 \) at the point \( P(1, 1) \) is \( y = 3x - 2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4dca08d-d7c1-4511-81d7-e3b17171b463%2Fe6e4884d-faf0-4fc9-be9d-388d19fe53c7%2Fe1cp25b_processed.png&w=3840&q=75)
Transcribed Image Text:**12. Find the equation of the tangent line to \( f(x) = x^3 \) at the point \( P(1, 1) \).**
To find the equation of the tangent line at a given point on a curve, we need to determine the slope of the tangent line and use the point-slope form of a line equation.
**Steps to solve:**
1. **Find the Derivative of the Function:**
The derivative \( f'(x) \) of \( f(x) = x^3 \) gives the slope of the tangent to the curve at any point \( x \).
\[
f'(x) = 3x^2
\]
2. **Calculate the Slope at the Specific Point:**
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \( P(1, 1) \).
\[
f'(1) = 3(1)^2 = 3
\]
3. **Use the Point-Slope Form of a Line:**
The point-slope form of the equation of a line is:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is a point on the line. Using \( m = 3 \) and \( (x_1, y_1) = (1, 1) \):
\[
y - 1 = 3(x - 1)
\]
4. **Simplify the Equation:**
Expand and rearrange the equation to standard form:
\[
y - 1 = 3x - 3
\]
\[
y = 3x - 2
\]
Therefore, the equation of the tangent line to \( f(x) = x^3 \) at the point \( P(1, 1) \) is \( y = 3x - 2 \).
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