Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 35E
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![**Example Problem: Finding Intervals of Increase**
**Problem Statement:**
Let \( f \) be the function given by
\[ f(x) = 300x - x^3. \]
On which of the following intervals is the function \( f \) increasing?
**Solution:**
To determine intervals where the function \( f \) is increasing, we need to find where its derivative, \( f'(x) \), is positive.
1. **Find the derivative \( f'(x) \):**
\[ f(x) = 300x - x^3 \]
Using the rules of differentiation:
\[ f'(x) = 300 - 3x^2 \]
2. **Set the derivative \( f'(x) \) equal to zero and solve for \( x \):**
\[ 300 - 3x^2 = 0 \]
\[ 3x^2 = 300 \]
\[ x^2 = 100 \]
\[ x = \pm 10 \]
3. **Determine the intervals to test:**
The critical points divide the x-axis into three intervals: \( (-\infty, -10) \), \( (-10, 10) \), and \( (10, \infty) \).
4. **Test points in each interval:**
- For \( x \) in \( (-\infty, -10) \), pick \( x = -11 \):
\[ f'(-11) = 300 - 3(-11)^2 = 300 - 3 \cdot 121 = 300 - 363 = -63 \]
Since \( f'(-11) < 0 \), \( f \) is decreasing on \( (-\infty, -10) \).
- For \( x \) in \( (-10, 10) \), pick \( x = 0 \):
\[ f'(0) = 300 - 3(0)^2 = 300 \]
Since \( f'(0) > 0 \), \( f \) is increasing on \( (-10, 10) \).
- For \( x \) in \( (10, \infty) \), pick \( x = 11 \):
\[ f'(11) = 300 - 3(11)^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc965ec1f-6ecd-467d-b5f1-022eb1dfa0a4%2F3afb150b-9ce2-4159-9a5e-1134dfefaf28%2Fc6otkt7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example Problem: Finding Intervals of Increase**
**Problem Statement:**
Let \( f \) be the function given by
\[ f(x) = 300x - x^3. \]
On which of the following intervals is the function \( f \) increasing?
**Solution:**
To determine intervals where the function \( f \) is increasing, we need to find where its derivative, \( f'(x) \), is positive.
1. **Find the derivative \( f'(x) \):**
\[ f(x) = 300x - x^3 \]
Using the rules of differentiation:
\[ f'(x) = 300 - 3x^2 \]
2. **Set the derivative \( f'(x) \) equal to zero and solve for \( x \):**
\[ 300 - 3x^2 = 0 \]
\[ 3x^2 = 300 \]
\[ x^2 = 100 \]
\[ x = \pm 10 \]
3. **Determine the intervals to test:**
The critical points divide the x-axis into three intervals: \( (-\infty, -10) \), \( (-10, 10) \), and \( (10, \infty) \).
4. **Test points in each interval:**
- For \( x \) in \( (-\infty, -10) \), pick \( x = -11 \):
\[ f'(-11) = 300 - 3(-11)^2 = 300 - 3 \cdot 121 = 300 - 363 = -63 \]
Since \( f'(-11) < 0 \), \( f \) is decreasing on \( (-\infty, -10) \).
- For \( x \) in \( (-10, 10) \), pick \( x = 0 \):
\[ f'(0) = 300 - 3(0)^2 = 300 \]
Since \( f'(0) > 0 \), \( f \) is increasing on \( (-10, 10) \).
- For \( x \) in \( (10, \infty) \), pick \( x = 11 \):
\[ f'(11) = 300 - 3(11)^2
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