Find an equation for the line tangent to the graph of f(x) = -5xe* at the point (2, f(2)). (Round your coefficients to three decimal places.) y =
Find an equation for the line tangent to the graph of f(x) = -5xe* at the point (2, f(2)). (Round your coefficients to three decimal places.) y =
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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![**Problem Statement:**
Find an equation for the line tangent to the graph of \( f(x) = -5xe^x \) at the point \( (2, f(2)) \). (Round your coefficients to three decimal places.)
**Solution:**
To find the equation of the tangent line, we need the slope of the tangent line at the given point, which requires finding the derivative of \( f(x) \).
1. **Differentiate \( f(x) = -5xe^x \).**
Using the product rule:
\[ \frac{d}{dx}[-5xe^x] = -5(e^x + xe^x) \]
Simplify the derivative:
\[ f'(x) = -5e^x - 5xe^x \]
2. **Evaluate the derivative at \( x = 2 \) to find the slope.**
\[ f'(2) = -5e^2 - 5 \cdot 2e^2 = -5e^2 - 10e^2 \]
\[ f'(2) = -15e^2 \]
3. **Calculate \( f(2) \) to find the point on the line.**
\[ f(2) = -5 \cdot 2e^2 = -10e^2 \]
4. **Use the point-slope form of a line:**
The point-slope form is:
\[ y - f(2) = f'(2)(x - 2) \]
Substituting the values:
\[ y + 10e^2 = -15e^2(x - 2) \]
5. **Simplify to the slope-intercept form \( y = mx + b \):**
Expand and simplify to get:
\[ y = -15e^2x + 30e^2 - 10e^2 \]
\[ y = -15e^2x + 20e^2 \]
**Final Answer:**
The equation of the tangent line is:
\[ y = -15e^2x + 20e^2 \]
The coefficients rounded to three decimal places are calculated using the value of \( e^2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe88a8cea-da71-40ab-aaed-a0e0afa0772e%2F6c45acf4-5542-45ec-ac3e-bb810c8a8a7d%2Fk8j6wvh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find an equation for the line tangent to the graph of \( f(x) = -5xe^x \) at the point \( (2, f(2)) \). (Round your coefficients to three decimal places.)
**Solution:**
To find the equation of the tangent line, we need the slope of the tangent line at the given point, which requires finding the derivative of \( f(x) \).
1. **Differentiate \( f(x) = -5xe^x \).**
Using the product rule:
\[ \frac{d}{dx}[-5xe^x] = -5(e^x + xe^x) \]
Simplify the derivative:
\[ f'(x) = -5e^x - 5xe^x \]
2. **Evaluate the derivative at \( x = 2 \) to find the slope.**
\[ f'(2) = -5e^2 - 5 \cdot 2e^2 = -5e^2 - 10e^2 \]
\[ f'(2) = -15e^2 \]
3. **Calculate \( f(2) \) to find the point on the line.**
\[ f(2) = -5 \cdot 2e^2 = -10e^2 \]
4. **Use the point-slope form of a line:**
The point-slope form is:
\[ y - f(2) = f'(2)(x - 2) \]
Substituting the values:
\[ y + 10e^2 = -15e^2(x - 2) \]
5. **Simplify to the slope-intercept form \( y = mx + b \):**
Expand and simplify to get:
\[ y = -15e^2x + 30e^2 - 10e^2 \]
\[ y = -15e^2x + 20e^2 \]
**Final Answer:**
The equation of the tangent line is:
\[ y = -15e^2x + 20e^2 \]
The coefficients rounded to three decimal places are calculated using the value of \( e^2 \).
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