Finding Absolute Extrema For F(x) = 8x^4 - 3ln(x^2) On [0.08, 15]
Introduction
In calculus, determining the absolute extrema of a function over a closed interval is a fundamental problem with numerous applications in optimization and other fields. The absolute extrema refer to the maximum and minimum values the function attains within the specified interval. This article delves into the process of finding the absolute extrema for the function f(x) = 8x^4 - 3ln(x^2) on the interval [0.08, 15]. We will employ calculus techniques, including finding critical points and evaluating the function at these points and the interval endpoints. The final answers will be presented in the form (x, f(x)), rounded to two decimal places.
Understanding Absolute Extrema
Before diving into the specific problem, let's clarify the concept of absolute extrema. The absolute maximum of a function on an interval is the largest value the function achieves within that interval, while the absolute minimum is the smallest value. These extreme values can occur at critical points (where the derivative is zero or undefined) or at the endpoints of the interval. The Extreme Value Theorem guarantees that a continuous function on a closed interval will have both an absolute maximum and an absolute minimum.
Steps to Find Absolute Extrema
To find the absolute extrema, we follow a systematic approach:
- Find the derivative of the function: This step is crucial for identifying critical points, where the function's slope is zero or undefined. These points are potential locations for local maxima or minima.
- Find the critical points: Critical points are the x-values where the derivative is equal to zero or undefined. These points are candidates for absolute extrema.
- Evaluate the function at the critical points and endpoints: Plug the critical points and the endpoints of the interval into the original function to find the corresponding y-values. This step helps determine the function's value at potential absolute extrema locations.
- Identify the absolute maximum and minimum: Compare the y-values obtained in the previous step. The largest y-value represents the absolute maximum, and the smallest y-value represents the absolute minimum on the interval.
Applying the Steps to f(x) = 8x^4 - 3ln(x^2) on [0.08, 15]
Let's apply these steps to our function, f(x) = 8x^4 - 3ln(x^2), on the interval [0.08, 15]. This detailed application will illustrate the process and provide a clear solution.
1. Find the Derivative of the Function
First, we need to find the derivative of f(x) with respect to x. Recall that the derivative of x^n is nx^(n-1) and the derivative of ln(u) is (1/u) * du/dx. Using the chain rule, we differentiate f(x) as follows:
f(x) = 8x^4 - 3ln(x^2)
f'(x) = d/dx (8x^4) - d/dx (3ln(x^2))
Applying the power rule and the chain rule:
f'(x) = 32x^3 - 3 * (1/x^2) * (2x)
f'(x) = 32x^3 - 6/x
2. Find the Critical Points
Critical points occur where the derivative is equal to zero or undefined. Let's set f'(x) = 0 and solve for x:
32x^3 - 6/x = 0
To solve this equation, we first eliminate the fraction by multiplying both sides by x:
32x^4 - 6 = 0
Now, we isolate x^4:
32x^4 = 6
x^4 = 6/32 = 3/16
Taking the fourth root of both sides:
x = ±(3/16)^(1/4)
x ≈ ±0.686
Since we are considering the interval [0.08, 15], we only take the positive root, x ≈ 0.686. Now, we need to check where the derivative is undefined. The derivative f'(x) = 32x^3 - 6/x is undefined when x = 0. However, x = 0 is not in our interval [0.08, 15], so we don't need to consider it. Thus, the only critical point within our interval is approximately x = 0.686.
3. Evaluate the Function at the Critical Points and Endpoints
Now, we evaluate the original function f(x) at the critical point x ≈ 0.686 and the endpoints of the interval, x = 0.08 and x = 15:
- At x = 0.08:
f(0.08) = 8(0.08)^4 - 3ln((0.08)^2)
f(0.08) ≈ 8(0.00004096) - 3ln(0.0064)
f(0.08) ≈ 0.00032768 - 3(-5.051)
f(0.08) ≈ 0.00032768 + 15.153
f(0.08) ≈ 15.15
- At x ≈ 0.686:
f(0.686) = 8(0.686)^4 - 3ln((0.686)^2)
f(0.686) ≈ 8(0.222) - 3ln(0.4706)
f(0.686) ≈ 1.776 - 3(-0.754)
f(0.686) ≈ 1.776 + 2.262
f(0.686) ≈ 4.04
- At x = 15:
f(15) = 8(15)^4 - 3ln((15)^2)
f(15) = 8(50625) - 3ln(225)
f(15) = 405000 - 3(5.416)
f(15) = 405000 - 16.248
f(15) ≈ 404983.75
4. Identify the Absolute Maximum and Minimum
Comparing the values of f(x) at the critical point and endpoints:
- f(0.08) ≈ 15.15
- f(0.686) ≈ 4.04
- f(15) ≈ 404983.75
The smallest value is approximately 4.04, which occurs at x ≈ 0.686, and the largest value is approximately 404983.75, which occurs at x = 15.
Results
Therefore, the absolute minimum is approximately (0.69, 4.04), and the absolute maximum is approximately (15, 404983.75). We have rounded the x-value of the absolute minimum to two decimal places as requested.
Conclusion
In conclusion, by following the steps of finding the derivative, identifying critical points, and evaluating the function at these points and the interval endpoints, we successfully determined the absolute extrema of the function f(x) = 8x^4 - 3ln(x^2) on the interval [0.08, 15]. The absolute minimum is approximately (0.69, 4.04), and the absolute maximum is approximately (15, 404983.75). This process highlights the importance of calculus in solving optimization problems and finding extreme values of functions.
