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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![**Question 3:** Determine where \( f(x) = 12x - x^3 \) has any local maximums or any local minimums.
**Analysis:**
To find the local maximums and minimums, we need to compute the first derivative of the function \( f(x) \) and then find the critical points by setting the derivative equal to zero.
1. **First Derivative:**
\[
f'(x) = \frac{d}{dx}(12x - x^3) = 12 - 3x^2
\]
2. **Critical Points:**
Set \( f'(x) = 0 \) to find critical points:
\[
12 - 3x^2 = 0
\]
\[
3x^2 = 12
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]
3. **Second Derivative Test:**
To determine whether these critical points are local maximums or minimums, use the second derivative:
\[
f''(x) = \frac{d}{dx}(-3x^2) = -6x
\]
- Evaluate \( f''(x) \) at the critical points:
- At \( x = 2 \):
\[
f''(2) = -6(2) = -12 \quad (\text{Negative, so } x = 2 \text{ is a local maximum})
\]
- At \( x = -2 \):
\[
f''(-2) = -6(-2) = 12 \quad (\text{Positive, so } x = -2 \text{ is a local minimum})
\]
**Conclusion:**
The function \( f(x) = 12x - x^3 \) has a local maximum at \( x = 2 \) and a local minimum at \( x = -2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb234898a-ca00-4740-ae3d-79576f672843%2F49569e81-9c40-43d8-a3eb-5c66e84acdcb%2F7ug14gm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 3:** Determine where \( f(x) = 12x - x^3 \) has any local maximums or any local minimums.
**Analysis:**
To find the local maximums and minimums, we need to compute the first derivative of the function \( f(x) \) and then find the critical points by setting the derivative equal to zero.
1. **First Derivative:**
\[
f'(x) = \frac{d}{dx}(12x - x^3) = 12 - 3x^2
\]
2. **Critical Points:**
Set \( f'(x) = 0 \) to find critical points:
\[
12 - 3x^2 = 0
\]
\[
3x^2 = 12
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]
3. **Second Derivative Test:**
To determine whether these critical points are local maximums or minimums, use the second derivative:
\[
f''(x) = \frac{d}{dx}(-3x^2) = -6x
\]
- Evaluate \( f''(x) \) at the critical points:
- At \( x = 2 \):
\[
f''(2) = -6(2) = -12 \quad (\text{Negative, so } x = 2 \text{ is a local maximum})
\]
- At \( x = -2 \):
\[
f''(-2) = -6(-2) = 12 \quad (\text{Positive, so } x = -2 \text{ is a local minimum})
\]
**Conclusion:**
The function \( f(x) = 12x - x^3 \) has a local maximum at \( x = 2 \) and a local minimum at \( x = -2 \).
Expert Solution

Step 1
To determine maxima or minima for the given function
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