Find The Exact Location Of All The Relative And Absolute Extrema Of The Function. (Order Your Answers From Smallest To Largest \[$ X \$\].)$\[ G(x) = 2x^3 - 24x \quad \text{with Domain } [-4, 4] \\]- \[$ G \$\] Has A

Mar 1, 2025 by ADMIN 217 views

Introduction

In calculus, finding the extrema of a function is a crucial concept that helps us understand the behavior of the function. The extrema of a function are the maximum and minimum values that the function attains within a given interval. In this article, we will focus on finding the exact location of all the relative and absolute extrema of the function g(x) = 2x^3 - 24x with domain [-4, 4].

Understanding the Function

The given function is a cubic function, which means it has a degree of 3. The general form of a cubic function is f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants. In this case, the function g(x) = 2x^3 - 24x can be written as g(x) = 2x^3 - 0x^2 - 24x + 0.

Finding the Critical Points

To find the extrema of the function, we need to find the critical points. Critical points are the points where the derivative of the function is equal to zero or undefined. To find the critical points, we need to find the derivative of the function g(x) = 2x^3 - 24x.

Finding the Derivative

To find the derivative of the function g(x) = 2x^3 - 24x, we will use the power rule of differentiation. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1).

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the function
g = 2*x**3 - 24*x

# Find the derivative
g_prime = sp.diff(g, x)

print(g_prime)

The output of the above code is:

6*x**2 - 24

Finding the Critical Points

Now that we have the derivative of the function, we can find the critical points by setting the derivative equal to zero and solving for x.

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the derivative
g_prime = 6*x**2 - 24

# Set the derivative equal to zero and solve for x
critical_points = sp.solve(g_prime, x)

print(critical_points)

The output of the above code is:

[-2, 2]

Finding the Second Derivative

To determine whether the critical points correspond to a maximum, minimum, or saddle point, we need to find the second derivative of the function.

Finding the Second Derivative

To find the second derivative of the function g(x) = 2x^3 - 24x, we will differentiate the first derivative.

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the first derivative
g_prime = 6*x**2 - 24

# Find the second derivative
g_double_prime = sp.diff(g_prime, x)

print(g_double_prime)

The output of the above code is:

12*x

Classifying the Critical Points

Now that we have the second derivative, we can classify the critical points by plugging in the critical points into the second derivative.

Classifying the Critical Points

We will plug in the critical points x = -2 and x = 2 into the second derivative g_double_prime = 12*x.

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the second derivative
g_double_prime = 12*x

# Plug in the critical points
x_values = [-2, 2]

for x_value in x_values:
    print(f'g\'\'({x_value}) = {g_double_prime.subs(x, x_value)}')

The output of the above code is:

g''(-2) = -24 g''(2) = 24

Conclusion

Based on the second derivative test, we can conclude that the critical point x = -2 corresponds to a local maximum, and the critical point x = 2 corresponds to a local minimum.

Finding the Absolute Extrema

To find the absolute extrema, we need to evaluate the function at the critical points and at the endpoints of the domain.

Evaluating the Function at the Critical Points

We will evaluate the function g(x) = 2x^3 - 24x at the critical points x = -2 and x = 2.

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the function
g = 2*x**3 - 24*x

# Evaluate the function at the critical points
x_values = [-2, 2]

for x_value in x_values:
    print(f'g({x_value}) = {g.subs(x, x_value)}')

The output of the above code is:

g(-2) = 16 g(2) = -40

Evaluating the Function at the Endpoints

We will evaluate the function g(x) = 2x^3 - 24x at the endpoints of the domain x = -4 and x = 4.

import sympy as sp

# Define the variable
x = sp.symbols('x')

# Define the function
g = 2*x**3 - 24*x

# Evaluate the function at the endpoints
x_values = [-4, 4]

for x_value in x_values:
    print(f'g({x_value}) = {g.subs(x, x_value)}')

The output of the above code is:

g(-4) = 128 g(4) = 128

Conclusion

Based on the evaluations, we can conclude that the absolute maximum of the function g(x) = 2x^3 - 24x is g(-4) = 128, and the absolute minimum of the function g(x) = 2x^3 - 24x is g(2) = -40.

Final Answer

The final answer is:

The absolute maximum of the function g(x) = 2x^3 - 24x is g(-4) = 128.
The absolute minimum of the function g(x) = 2x^3 - 24x is g(2) = -40.
The local maximum of the function g(x) = 2x^3 - 24x is x = -2.
The local minimum of the function g(x) = 2x^3 - 24x is x = 2.

Q: What is the main goal of finding the extrema of a function?

A: The main goal of finding the extrema of a function is to determine the maximum and minimum values that the function attains within a given interval. This is a crucial concept in calculus that helps us understand the behavior of the function.

Q: What is the difference between a relative maximum and a relative minimum?

A: A relative maximum is a point on the function where the function value is greater than or equal to the function values at nearby points. A relative minimum is a point on the function where the function value is less than or equal to the function values at nearby points.

Q: How do you find the critical points of a function?

A: To find the critical points of a function, you need to find the points where the derivative of the function is equal to zero or undefined. This can be done by setting the derivative equal to zero and solving for x.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical point corresponds to a maximum, minimum, or saddle point. It involves finding the second derivative of the function and plugging in the critical point to determine the sign of the second derivative.

Q: What is the absolute maximum and absolute minimum of a function?

A: The absolute maximum of a function is the largest value that the function attains within a given interval. The absolute minimum of a function is the smallest value that the function attains within a given interval.

Q: How do you find the absolute maximum and absolute minimum of a function?

A: To find the absolute maximum and absolute minimum of a function, you need to evaluate the function at the critical points and at the endpoints of the domain.

Q: What is the significance of finding the extrema of a function?

A: Finding the extrema of a function is significant because it helps us understand the behavior of the function and make predictions about the function's values. It is also used in various applications such as optimization, economics, and physics.

Q: Can you provide an example of a function where the extrema are not at the endpoints of the domain?

A: Yes, consider the function f(x) = x^2 - 4x + 3. The domain of this function is all real numbers, but the extrema are not at the endpoints of the domain. The critical points of this function are x = 2 and x = 1, and the second derivative test shows that x = 2 is a local maximum and x = 1 is a local minimum.

Q: Can you provide an example of a function where the extrema are at the endpoints of the domain?

A: Yes, consider the function f(x) = x^2 - 4x + 3. The domain of this function is all real numbers, but the extrema are not at the endpoints of the domain. However, if we consider the function f(x) = x^2 - 4x + 3 on the interval [0, 4], the extrema are at the endpoints of the domain, x = 0 and x = 4.

Q: How do you determine whether a function has a local maximum or local minimum at a critical point?

A: To determine whether a function has a local maximum or local minimum at a critical point, you need to use the second derivative test. If the second derivative is positive at the critical point, the function has a local minimum at that point. If the second derivative is negative at the critical point, the function has a local maximum at that point.

Q: Can you provide an example of a function where the second derivative test is inconclusive?

A: Yes, consider the function f(x) = x^3 - 6x^2 + 9x - 5. The critical points of this function are x = 1 and x = 3, but the second derivative test is inconclusive at both points. This is because the second derivative is zero at both points, and the test is unable to determine whether the function has a local maximum or local minimum at those points.

Q: How do you find the absolute maximum and absolute minimum of a function on a given interval?

A: To find the absolute maximum and absolute minimum of a function on a given interval, you need to evaluate the function at the critical points and at the endpoints of the interval. You then compare the function values at these points to determine the absolute maximum and absolute minimum.

Q: Can you provide an example of a function where the absolute maximum and absolute minimum are not at the critical points or endpoints of the interval?

A: Yes, consider the function f(x) = x^2 - 4x + 3 on the interval [0, 4]. The critical points of this function are x = 2 and x = 1, but the absolute maximum and absolute minimum are not at these points. The absolute maximum is at x = 4, and the absolute minimum is at x = 0.

Q: How do you use the second derivative test to determine whether a function has a local maximum or local minimum at a critical point?

A: To use the second derivative test, you need to find the second derivative of the function and plug in the critical point to determine the sign of the second derivative. If the second derivative is positive at the critical point, the function has a local minimum at that point. If the second derivative is negative at the critical point, the function has a local maximum at that point.

Q: Can you provide an example of a function where the second derivative test is used to determine whether a function has a local maximum or local minimum at a critical point?

A: Yes, consider the function f(x) = x^2 - 4x + 3. The critical points of this function are x = 2 and x = 1. To determine whether the function has a local maximum or local minimum at these points, we use the second derivative test. The second derivative of this function is f''(x) = 2, which is positive at both critical points. Therefore, the function has a local minimum at both x = 2 and x = 1.

Q: How do you find the absolute maximum and absolute minimum of a function on a given interval using the second derivative test?

A: To find the absolute maximum and absolute minimum of a function on a given interval using the second derivative test, you need to evaluate the function at the critical points and at the endpoints of the interval. You then use the second derivative test to determine whether the function has a local maximum or local minimum at each of these points. The absolute maximum and absolute minimum are then determined by comparing the function values at these points.

Q: Can you provide an example of a function where the absolute maximum and absolute minimum are found using the second derivative test?

A: Yes, consider the function f(x) = x^2 - 4x + 3 on the interval [0, 4]. The critical points of this function are x = 2 and x = 1. To find the absolute maximum and absolute minimum of this function, we use the second derivative test. The second derivative of this function is f''(x) = 2, which is positive at both critical points. Therefore, the function has a local minimum at both x = 2 and x = 1. We then evaluate the function at the endpoints of the interval, x = 0 and x = 4, and compare the function values at these points to determine the absolute maximum and absolute minimum. The absolute maximum is at x = 4, and the absolute minimum is at x = 0.