Answer :
To solve the given problem, we need to analyze the potential energy function [tex]\( U(x) \)[/tex] and find its critical points, as well as determine the nature of these critical points to identify the maximum and minimum potential energy values. Let's go through this step-by-step.
1. Given Potential Energy Function:
[tex]\[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
2. Find the First Derivative:
We need to find the first derivative [tex]\( U'(x) \)[/tex] to locate the critical points:
[tex]\[ U'(x) = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U'(x) = x^3 - x \][/tex]
3. Set the First Derivative to Zero:
To find the critical points, we set [tex]\( U'(x) = 0 \)[/tex]:
[tex]\[ x^3 - x = 0 \][/tex]
Factorizing the equation:
[tex]\[ x(x^2 - 1) = 0 \][/tex]
[tex]\[ x(x-1)(x+1) = 0 \][/tex]
This gives us the critical points:
[tex]\[ x = -1, 0, 1 \][/tex]
4. Evaluate the Potential Energy at the Critical Points:
Next, we substitute each critical point back into the potential energy function to find their corresponding values:
[tex]\[ U(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
[tex]\[ U(0) = \frac{0^4}{4} - \frac{0^2}{2} = 0 \][/tex]
[tex]\[ U(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
Therefore, the potential energy values at the critical points are:
[tex]\[ U(-1) = -\frac{1}{4}, \quad U(0) = 0, \quad U(1) = -\frac{1}{4} \][/tex]
5. Find the Second Derivative:
We need the second derivative [tex]\( U''(x) \)[/tex] to determine if each critical point is a maximum or minimum:
[tex]\[ U''(x) = \frac{d^2}{dx^2}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U''(x) = 3x^2 - 1 \][/tex]
6. Evaluate the Second Derivative at the Critical Points:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ U''(-1) = 3(-1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ U''(0) = 3(0)^2 - 1 = -1 \quad (\text{negative, maximum}) \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ U''(1) = 3(1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
7. Identify the Maximum and Minimum Potential Energy:
- The maximum potential energy is at [tex]\( x = 0 \)[/tex] with [tex]\( U(0) = 0 \)[/tex]
- The minimum potential energy is at [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex] with [tex]\( U(-1) = U(1) = -\frac{1}{4} \)[/tex]
Thus, the maximum potential energy is [tex]\( \boxed{0} \)[/tex] and the minimum potential energy is [tex]\( \boxed{-\frac{1}{4}} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(3) \ 0, -\frac{1}{4}} \][/tex]
1. Given Potential Energy Function:
[tex]\[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
2. Find the First Derivative:
We need to find the first derivative [tex]\( U'(x) \)[/tex] to locate the critical points:
[tex]\[ U'(x) = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U'(x) = x^3 - x \][/tex]
3. Set the First Derivative to Zero:
To find the critical points, we set [tex]\( U'(x) = 0 \)[/tex]:
[tex]\[ x^3 - x = 0 \][/tex]
Factorizing the equation:
[tex]\[ x(x^2 - 1) = 0 \][/tex]
[tex]\[ x(x-1)(x+1) = 0 \][/tex]
This gives us the critical points:
[tex]\[ x = -1, 0, 1 \][/tex]
4. Evaluate the Potential Energy at the Critical Points:
Next, we substitute each critical point back into the potential energy function to find their corresponding values:
[tex]\[ U(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
[tex]\[ U(0) = \frac{0^4}{4} - \frac{0^2}{2} = 0 \][/tex]
[tex]\[ U(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
Therefore, the potential energy values at the critical points are:
[tex]\[ U(-1) = -\frac{1}{4}, \quad U(0) = 0, \quad U(1) = -\frac{1}{4} \][/tex]
5. Find the Second Derivative:
We need the second derivative [tex]\( U''(x) \)[/tex] to determine if each critical point is a maximum or minimum:
[tex]\[ U''(x) = \frac{d^2}{dx^2}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U''(x) = 3x^2 - 1 \][/tex]
6. Evaluate the Second Derivative at the Critical Points:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ U''(-1) = 3(-1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ U''(0) = 3(0)^2 - 1 = -1 \quad (\text{negative, maximum}) \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ U''(1) = 3(1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
7. Identify the Maximum and Minimum Potential Energy:
- The maximum potential energy is at [tex]\( x = 0 \)[/tex] with [tex]\( U(0) = 0 \)[/tex]
- The minimum potential energy is at [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex] with [tex]\( U(-1) = U(1) = -\frac{1}{4} \)[/tex]
Thus, the maximum potential energy is [tex]\( \boxed{0} \)[/tex] and the minimum potential energy is [tex]\( \boxed{-\frac{1}{4}} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(3) \ 0, -\frac{1}{4}} \][/tex]
