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 the equation of the tangent line to the graph of \( f(x) = 7 \sin(x) + 6 \cos(x) \) at \( x = \frac{\pi}{2} \) in the form \( y = mx + b \).
**Solution:**
To find the equation of the tangent line, we need to determine the slope of the tangent line at \( x = \frac{\pi}{2} \) and the point on the graph \( (x, f(x)) \).
1. **Calculate \( f(x) \) at \( x = \frac{\pi}{2} \):**
\[
f\left(\frac{\pi}{2}\right) = 7 \sin\left(\frac{\pi}{2}\right) + 6 \cos\left(\frac{\pi}{2}\right)
\]
\[
f\left(\frac{\pi}{2}\right) = 7(1) + 6(0) = 7
\]
Thus, the point on the graph is \( \left(\frac{\pi}{2}, 7\right) \).
2. **Calculate the derivative \( f'(x) \):**
\[
f'(x) = \frac{d}{dx}[7 \sin(x) + 6 \cos(x)] = 7 \cos(x) - 6 \sin(x)
\]
3. **Determine the slope at \( x = \frac{\pi}{2} \):**
\[
f'\left(\frac{\pi}{2}\right) = 7 \cos\left(\frac{\pi}{2}\right) - 6 \sin\left(\frac{\pi}{2}\right)
\]
\[
f'\left(\frac{\pi}{2}\right) = 7 (0) - 6 (1) = -6
\]
4. **Equation of the tangent line:**
Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = \left(\frac{\pi}{2}, 7\right) \) and \( m = -6 \):
\[
y - 7 = -6 \left(x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c89e543-4ffd-4c36-8dc6-0952cfcad04e%2F888cacf5-5280-4d8b-93eb-e8de15cd5293%2F7jmp4ih_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the equation of the tangent line to the graph of \( f(x) = 7 \sin(x) + 6 \cos(x) \) at \( x = \frac{\pi}{2} \) in the form \( y = mx + b \).
**Solution:**
To find the equation of the tangent line, we need to determine the slope of the tangent line at \( x = \frac{\pi}{2} \) and the point on the graph \( (x, f(x)) \).
1. **Calculate \( f(x) \) at \( x = \frac{\pi}{2} \):**
\[
f\left(\frac{\pi}{2}\right) = 7 \sin\left(\frac{\pi}{2}\right) + 6 \cos\left(\frac{\pi}{2}\right)
\]
\[
f\left(\frac{\pi}{2}\right) = 7(1) + 6(0) = 7
\]
Thus, the point on the graph is \( \left(\frac{\pi}{2}, 7\right) \).
2. **Calculate the derivative \( f'(x) \):**
\[
f'(x) = \frac{d}{dx}[7 \sin(x) + 6 \cos(x)] = 7 \cos(x) - 6 \sin(x)
\]
3. **Determine the slope at \( x = \frac{\pi}{2} \):**
\[
f'\left(\frac{\pi}{2}\right) = 7 \cos\left(\frac{\pi}{2}\right) - 6 \sin\left(\frac{\pi}{2}\right)
\]
\[
f'\left(\frac{\pi}{2}\right) = 7 (0) - 6 (1) = -6
\]
4. **Equation of the tangent line:**
Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = \left(\frac{\pi}{2}, 7\right) \) and \( m = -6 \):
\[
y - 7 = -6 \left(x
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