Consider the function f(x) = x - 72a + 9, – 5 < x < 13. This function has an absolute minimum value equal to and an absolute maximum value equal to
Consider the function f(x) = x - 72a + 9, – 5 < x < 13. This function has an absolute minimum value equal to and an absolute maximum value equal to
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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Consider the function f(x) = x^4 - 72x^2 + 9, -5
![### Example Problem on Finding Absolute Minimum and Maximum Values
**Problem Statement:**
Consider the function \( f(x) = x^4 - 72x^2 + 9 \), for the domain \(-5 \leq x \leq 13\).
1. This function has an absolute minimum value equal to ______
2. and an absolute maximum value equal to ______
**Solution Approach:**
To find the absolute minimum and maximum values of \( f(x) \) in the given interval, follow these steps:
1. **Find the critical points** within the interval \([-5, 13]\) by setting the derivative \( f'(x) \) to zero and solving for \( x \).
2. **Evaluate the function** at the critical points and at the endpoints of the interval.
3. **Compare** the function values to determine the absolute minimum and maximum.
**Compute Derivative:**
Find \( f'(x) \):
\[ f(x) = x^4 - 72x^2 + 9 \]
\[ f'(x) = 4x^3 - 144x \]
Set the derivative equal to zero to find critical points:
\[ 4x^3 - 144x = 0 \]
Factor out common terms:
\[ 4x(x^2 - 36) = 0 \]
\[ 4x(x - 6)(x + 6) = 0 \]
Thus, the critical points are:
\[ x = 0, x = 6, x = -6 \]
**Evaluate \( f(x) \) at the critical points and endpoints:**
For \( x = -6 \):
\[ f(-6) = (-6)^4 - 72(-6)^2 + 9 = 1296 - 2592 + 9 = -1287 \]
For \( x = 0 \):
\[ f(0) = 0^4 - 72(0)^2 + 9 = 9 \]
For \( x = 6 \):
\[ f(6) = 6^4 - 72(6)^2 + 9 = 1296 - 2592 + 9 = -1287 \]
For \( x = -5 \):
\[ f(-5) = (-5)^4 - 72(-5)^2 + 9 = 625 - 1800 + 9 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0d58f0b-90fc-4a9f-a40d-3617c86d0254%2Ff003f40a-bd28-4bd6-99a4-cce3f625a496%2Fkm5k15q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem on Finding Absolute Minimum and Maximum Values
**Problem Statement:**
Consider the function \( f(x) = x^4 - 72x^2 + 9 \), for the domain \(-5 \leq x \leq 13\).
1. This function has an absolute minimum value equal to ______
2. and an absolute maximum value equal to ______
**Solution Approach:**
To find the absolute minimum and maximum values of \( f(x) \) in the given interval, follow these steps:
1. **Find the critical points** within the interval \([-5, 13]\) by setting the derivative \( f'(x) \) to zero and solving for \( x \).
2. **Evaluate the function** at the critical points and at the endpoints of the interval.
3. **Compare** the function values to determine the absolute minimum and maximum.
**Compute Derivative:**
Find \( f'(x) \):
\[ f(x) = x^4 - 72x^2 + 9 \]
\[ f'(x) = 4x^3 - 144x \]
Set the derivative equal to zero to find critical points:
\[ 4x^3 - 144x = 0 \]
Factor out common terms:
\[ 4x(x^2 - 36) = 0 \]
\[ 4x(x - 6)(x + 6) = 0 \]
Thus, the critical points are:
\[ x = 0, x = 6, x = -6 \]
**Evaluate \( f(x) \) at the critical points and endpoints:**
For \( x = -6 \):
\[ f(-6) = (-6)^4 - 72(-6)^2 + 9 = 1296 - 2592 + 9 = -1287 \]
For \( x = 0 \):
\[ f(0) = 0^4 - 72(0)^2 + 9 = 9 \]
For \( x = 6 \):
\[ f(6) = 6^4 - 72(6)^2 + 9 = 1296 - 2592 + 9 = -1287 \]
For \( x = -5 \):
\[ f(-5) = (-5)^4 - 72(-5)^2 + 9 = 625 - 1800 + 9 =
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