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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Q7. Please answer this question
![**Problem Statement:**
Find an equation of the tangent line to the curve at the given point.
**Function:**
\[ y = 2x^3 - x^2 + 1 \]
**Given Point:**
\((2, 13)\)
**Solution:**
To solve this, we need to find the derivative of the function to get the slope of the tangent line at the given point. Then, use the point-slope form to find the equation.
**Derivative:**
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3 - x^2 + 1) \]
\[ = 6x^2 - 2x \]
**Slope at \(x = 2\):**
\[ m = 6(2)^2 - 2(2) \]
\[ = 6(4) - 4 \]
\[ = 24 - 4 \]
\[ = 20 \]
**Equation of the Tangent Line:**
Using the point-slope form: \( y - y_1 = m(x - x_1) \)
\[ y - 13 = 20(x - 2) \]
\[ y - 13 = 20x - 40 \]
\[ y = 20x - 27 \]
**Final Answer:**
\[ y = 20x - 27 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59d94971-c4ee-4188-b521-ed6cff9deaa9%2Fba189a42-4308-4a9d-8e5f-6e94a11e8b95%2Ftyhac8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation of the tangent line to the curve at the given point.
**Function:**
\[ y = 2x^3 - x^2 + 1 \]
**Given Point:**
\((2, 13)\)
**Solution:**
To solve this, we need to find the derivative of the function to get the slope of the tangent line at the given point. Then, use the point-slope form to find the equation.
**Derivative:**
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3 - x^2 + 1) \]
\[ = 6x^2 - 2x \]
**Slope at \(x = 2\):**
\[ m = 6(2)^2 - 2(2) \]
\[ = 6(4) - 4 \]
\[ = 24 - 4 \]
\[ = 20 \]
**Equation of the Tangent Line:**
Using the point-slope form: \( y - y_1 = m(x - x_1) \)
\[ y - 13 = 20(x - 2) \]
\[ y - 13 = 20x - 40 \]
\[ y = 20x - 27 \]
**Final Answer:**
\[ y = 20x - 27 \]
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