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 1:**
Find all the values of \( x \) where the function \( f(x) = 5e^x \sin x \) has a horizontal tangent line.
**Explanation:**
To determine where the function has a horizontal tangent line, we need to find where the derivative \( f'(x) \) equals zero.
1. **Differentiate the function:** Use the product rule for differentiation, as the function is a product of \( 5e^x \) and \( \sin x \).
The product rule states: \( (uv)' = u'v + uv' \).
Let \( u = 5e^x \) and \( v = \sin x \).
- \( u' = 5e^x \)
- \( v' = \cos x \)
Therefore, \( f'(x) = (5e^x)(\cos x) + (5e^x)(\sin x) \).
2. **Set the derivative to zero:**
\[ 5e^x ( \cos x + \sin x ) = 0 \]
Since \( 5e^x \neq 0 \) for any real \( x \), we focus on:
\[ \cos x + \sin x = 0 \]
3. **Solve for \( x \):**
This simplifies to:
\[ \sin x = -\cos x \]
\[ \tan x = -1 \]
The solutions to \( \tan x = -1 \) are:
\[ x = \frac{3\pi}{4} + n\pi \]
for any integer \( n \).
These steps are critical for someone learning how to find where a function has a horizontal tangent line, which is where the slope of the tangent line is zero.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F310a3344-537a-48c1-a1a5-be9fa2e55789%2F80ae1ff8-e6fb-48af-a959-577e8761931b%2Fu241tyh_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 1:**
Find all the values of \( x \) where the function \( f(x) = 5e^x \sin x \) has a horizontal tangent line.
**Explanation:**
To determine where the function has a horizontal tangent line, we need to find where the derivative \( f'(x) \) equals zero.
1. **Differentiate the function:** Use the product rule for differentiation, as the function is a product of \( 5e^x \) and \( \sin x \).
The product rule states: \( (uv)' = u'v + uv' \).
Let \( u = 5e^x \) and \( v = \sin x \).
- \( u' = 5e^x \)
- \( v' = \cos x \)
Therefore, \( f'(x) = (5e^x)(\cos x) + (5e^x)(\sin x) \).
2. **Set the derivative to zero:**
\[ 5e^x ( \cos x + \sin x ) = 0 \]
Since \( 5e^x \neq 0 \) for any real \( x \), we focus on:
\[ \cos x + \sin x = 0 \]
3. **Solve for \( x \):**
This simplifies to:
\[ \sin x = -\cos x \]
\[ \tan x = -1 \]
The solutions to \( \tan x = -1 \) are:
\[ x = \frac{3\pi}{4} + n\pi \]
for any integer \( n \).
These steps are critical for someone learning how to find where a function has a horizontal tangent line, which is where the slope of the tangent line is zero.
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