Finding The Minimum Value Of F(x) = 3x² - √(8x) + 32

Introduction: Understanding the Function and the Problem

In this article, we delve into the realm of mathematical functions to explore a fascinating problem: finding the least value of a given function. Specifically, we will be analyzing the function f(x) = 3x² - √(8x) + 32. This task involves a blend of algebraic manipulation, calculus concepts, and a keen understanding of function behavior. To effectively tackle this problem, we'll embark on a step-by-step journey, starting with a preliminary analysis of the function's domain and characteristics. We will then explore various methods to identify potential minimum points, such as utilizing derivatives and examining the function's graph. Our goal is not only to find the minimum value but also to provide a comprehensive explanation of the underlying principles and techniques involved. This exploration will not only enhance your problem-solving skills but also deepen your understanding of mathematical functions and their properties. By the end of this article, you will have a clear understanding of how to approach similar optimization problems, equipping you with valuable tools for tackling mathematical challenges in various contexts.

Our journey begins by acknowledging that this function is a combination of a quadratic term (3x²) and a square root term (-√(8x)), further augmented by a constant (32). This unique structure necessitates a careful approach to finding its minimum value. We can't simply apply standard quadratic minimization techniques because of the presence of the square root term. Therefore, we need to consider the interplay between these terms and how they influence the function's overall behavior. As we proceed, we will emphasize the importance of understanding the function's domain, as the square root term imposes a restriction on the possible values of x. This constraint will play a crucial role in determining the location of the minimum value. Furthermore, we will explore graphical representations of the function to gain visual insights into its shape and potential minimum points. By combining analytical techniques with visual aids, we aim to provide a holistic and intuitive understanding of the problem.

Ultimately, finding the minimum value of a function like f(x) = 3x² - √(8x) + 32 is not just an academic exercise. It has practical applications in various fields, including optimization problems in engineering, economics, and computer science. For instance, engineers might use similar techniques to minimize the cost of a structure while maintaining its strength, or economists might seek to minimize production costs while maximizing profits. Therefore, mastering these concepts and techniques is not only valuable for mathematical pursuits but also for solving real-world problems. Our exploration will not only focus on the specific function at hand but also highlight the broader applicability of the methods we employ. By understanding the underlying principles, you will be able to adapt these techniques to a wide range of optimization challenges. So, let's embark on this journey of mathematical discovery, where we unravel the secrets of this function and learn how to find its least value.

1. Determining the Domain of the Function

The domain of a function is the set of all possible input values (x-values) for which the function produces a real output. In the case of f(x) = 3x² - √(8x) + 32, the presence of the square root term, √(8x), imposes a restriction. The expression inside the square root, 8x, must be greater than or equal to zero to avoid imaginary numbers. This constraint leads us to the inequality 8x ≥ 0. Solving this inequality, we find that x ≥ 0. Therefore, the domain of the function f(x) is all non-negative real numbers, often expressed as [0, ∞). Understanding the domain is crucial because it limits the range of values we need to consider when searching for the minimum value. We only need to focus on x-values within the domain, as values outside this range are not valid inputs for the function.

The determination of the domain is a fundamental step in analyzing any function, especially those involving radicals, rational expressions, or logarithms. Each of these types of expressions can introduce restrictions on the possible input values. For example, a rational expression (a fraction with x in the denominator) would have a domain excluding any x-values that make the denominator zero. Similarly, a logarithmic function has a domain restricted to positive values, as the logarithm of a non-positive number is undefined in the real number system. By carefully identifying and addressing these restrictions, we ensure that our analysis is mathematically sound and that we are only considering valid inputs. In the context of our problem, the domain restriction significantly simplifies the search for the minimum value. It allows us to focus our attention on the behavior of the function within the interval [0, ∞), avoiding the need to consider negative x-values that would make the square root term imaginary.

Furthermore, understanding the domain can provide valuable insights into the overall behavior of the function. It can help us identify potential discontinuities, asymptotes, or other critical points that might influence the location of the minimum value. For instance, if the function had a discontinuity within its domain, we would need to carefully examine the function's behavior near that discontinuity to ensure that we are not overlooking a potential minimum. In our case, since the domain is [0, ∞), we know that the function is defined at x = 0, which means we need to consider the function's value at the boundary of its domain. This is a common consideration in optimization problems, as the minimum value can sometimes occur at the endpoints of the domain. By meticulously determining the domain and considering its implications, we lay a solid foundation for our subsequent analysis, ensuring that we are on the right track to finding the least value of the function.

2. Finding the Derivative of the Function

To find the minimum value of the function f(x) = 3x² - √(8x) + 32, we can employ the powerful tool of calculus: derivatives. The derivative of a function, denoted as f'(x), represents the instantaneous rate of change of the function with respect to x. At a minimum point, the function's rate of change is zero (or undefined), indicating a potential location for the minimum value. Therefore, finding the derivative and setting it equal to zero is a crucial step in locating these critical points.

To calculate the derivative, we'll apply the rules of differentiation to each term in the function. The derivative of 3x² is 6x, using the power rule (d/dx(x^n) = nx^(n-1)). For the square root term, -√(8x), we can rewrite it as -√(8) * x^(1/2). Applying the power rule again, the derivative of -√(8) * x^(1/2) is -√(8) * (1/2) * x^(-1/2), which simplifies to -√(2) / √x. The derivative of the constant term, 32, is zero. Therefore, the derivative of the function f(x) is f'(x) = 6x - √(2) / √x. This derivative represents the slope of the tangent line to the function at any given point x. By analyzing this derivative, we can identify the points where the function's slope is zero, which are potential locations for minimum or maximum values.

The process of differentiation is a cornerstone of calculus and has wide-ranging applications in various fields. It allows us to analyze the behavior of functions, such as their increasing or decreasing intervals, concavity, and points of inflection. In optimization problems, finding the derivative is essential for locating critical points, which are potential locations for minima or maxima. However, it's important to note that setting the derivative equal to zero only identifies potential extrema. We still need to verify whether these critical points correspond to actual minimum or maximum values using further analysis, such as the second derivative test or by examining the function's behavior around these points. Furthermore, it's crucial to be mindful of the domain of the function when interpreting the results of differentiation. Critical points that lie outside the domain are not valid candidates for extrema. In our case, since the domain of f(x) is [0, ∞), we only need to consider critical points within this interval. By carefully finding the derivative and considering its implications, we take a significant step towards finding the least value of the function.

3. Setting the Derivative to Zero and Solving for Critical Points

After finding the derivative of the function f(x), which is f'(x) = 6x - √(2) / √x, the next crucial step is to identify the critical points. These are the points where the derivative is either equal to zero or undefined. Setting the derivative to zero allows us to find potential locations for minimum or maximum values of the function. To solve for the critical points, we set f'(x) = 0, which gives us the equation 6x - √(2) / √x = 0. To solve this equation, we first isolate the term with the square root: 6x = √(2) / √x. Next, we multiply both sides by √x to get rid of the fraction: 6x√x = √(2). We can rewrite x√x as x^(3/2), so the equation becomes 6x^(3/2) = √(2). Now, we divide both sides by 6: x^(3/2) = √(2) / 6. To solve for x, we raise both sides to the power of 2/3: x = (√(2) / 6)^(2/3).

This algebraic manipulation is a common technique used in solving equations involving radicals and exponents. By carefully isolating terms and applying inverse operations, we can isolate the variable and find its value. In this case, we have found a potential critical point. However, it's important to note that we need to verify whether this critical point actually corresponds to a minimum or maximum value. We also need to check if there are any other critical points where the derivative is undefined. The derivative f'(x) = 6x - √(2) / √x is undefined when √x = 0, which occurs at x = 0. Therefore, x = 0 is another critical point that we need to consider. These critical points are the key to finding the minimum value of the function. They represent the points where the function's slope changes direction, potentially indicating a minimum or maximum. By identifying these points, we narrow down the possibilities and can focus our analysis on these specific locations.

The identification of critical points is a fundamental step in optimization problems. However, it's not the end of the process. We still need to determine whether these points correspond to minima, maxima, or neither. This can be done using various techniques, such as the second derivative test or by examining the function's behavior around the critical points. Furthermore, it's crucial to remember the domain of the function. Any critical points that lie outside the domain are not valid candidates for extrema. In our case, both critical points, x = 0 and x = (√(2) / 6)^(2/3), are within the domain [0, ∞), so we need to consider both of them. By carefully finding the critical points and understanding their significance, we are well on our way to finding the least value of the function.

4. Evaluating the Function at Critical Points and Endpoints

Having identified the critical points x = 0 and x = (√(2) / 6)^(2/3), the next step is to evaluate the function f(x) = 3x² - √(8x) + 32 at these points. This will give us the function's value at these potential minimum or maximum locations. Additionally, since the domain of the function is [0, ∞), we also need to consider the behavior of the function as x approaches infinity. However, since we are looking for the minimum value, and the 3x² term will dominate as x becomes very large, we can infer that the minimum value is unlikely to occur as x approaches infinity. Therefore, we will focus on evaluating the function at the critical points.

First, let's evaluate f(x) at x = 0: f(0) = 3(0)² - √(8(0)) + 32 = 0 - 0 + 32 = 32. This gives us a potential minimum value of 32. Next, we need to evaluate f(x) at x = (√(2) / 6)^(2/3). This requires some calculation. Let's denote this critical point as x*. Then x* = (√(2) / 6)^(2/3) ≈ 0.211. Now we substitute this value into the function: f(x*) = 3(0.211)² - √(8(0.211)) + 32 ≈ 3(0.0445) - √(1.688) + 32 ≈ 0.1335 - 1.299 + 32 ≈ 30.8345. Comparing the function values at the critical points, we have f(0) = 32 and f(x*) ≈ 30.8345. This suggests that the minimum value occurs at x = (√(2) / 6)^(2/3), and the minimum value is approximately 30.8345.

The evaluation of the function at critical points and endpoints is a crucial step in optimization problems. It allows us to compare the function's values at different locations and identify the absolute minimum or maximum. In our case, by evaluating the function at the critical points, we have narrowed down the possible locations for the minimum value. However, it's important to note that this is just a numerical approximation. To confirm that this is indeed the minimum value, we could use the second derivative test or analyze the function's behavior around the critical points. The second derivative test involves finding the second derivative of the function and evaluating it at the critical points. If the second derivative is positive at a critical point, it indicates a local minimum. If it's negative, it indicates a local maximum. If it's zero, the test is inconclusive. By carefully evaluating the function and considering the results of the second derivative test (if applicable), we can confidently determine the least value of the function.

5. Verifying the Minimum Value (Optional)

While we have found a potential minimum value by evaluating the function at critical points, it's always a good practice to verify our result. One common method for verification is the second derivative test. The second derivative of a function, denoted as f''(x), represents the rate of change of the first derivative. If the second derivative is positive at a critical point, it indicates that the function is concave up at that point, suggesting a local minimum. If the second derivative is negative, it indicates a local maximum. If the second derivative is zero, the test is inconclusive.

To apply the second derivative test, we first need to find the second derivative of our function f(x) = 3x² - √(8x) + 32. We already found the first derivative, f'(x) = 6x - √(2) / √x. To find the second derivative, we differentiate f'(x) with respect to x. The derivative of 6x is 6. To differentiate -√(2) / √x, we can rewrite it as -√(2) * x^(-1/2). Applying the power rule, the derivative of -√(2) * x^(-1/2) is -√(2) * (-1/2) * x^(-3/2), which simplifies to √(2) / (2x^(3/2)). Therefore, the second derivative is f''(x) = 6 + √(2) / (2x^(3/2)).

Now we need to evaluate the second derivative at the critical point x* = (√(2) / 6)^(2/3). Since x* is positive, the term √(2) / (2x*^(3/2)) is also positive. Therefore, f''(x*) = 6 + √(2) / (2x*^(3/2)) is positive. This confirms that the function is concave up at x*, indicating a local minimum. We can also observe that f''(x) is positive for all x in the domain [0, ∞), which means the function is concave up throughout its domain. This further supports the conclusion that x* is the location of the global minimum. By performing the second derivative test, we have strengthened our confidence in the result obtained in the previous step. While this step is optional, it provides a rigorous verification of the minimum value and ensures that our solution is mathematically sound.

Conclusion: The Least Value of the Function

In this comprehensive analysis, we have successfully determined the least value of the function f(x) = 3x² - √(8x) + 32. Our journey began with understanding the function's domain, which is crucial for any optimization problem. We identified that the domain is [0, ∞) due to the presence of the square root term. Next, we found the derivative of the function, f'(x) = 6x - √(2) / √x, which represents the instantaneous rate of change of the function. By setting the derivative to zero, we identified the critical points, which are potential locations for minimum or maximum values. We found two critical points: x = 0 and x = (√(2) / 6)^(2/3).

We then evaluated the function at the critical points to determine the function's value at these locations. We found that f(0) = 32 and f((√(2) / 6)^(2/3)) ≈ 30.8345. Comparing these values, we concluded that the minimum value occurs at x = (√(2) / 6)^(2/3), and the minimum value is approximately 30.8345. To further strengthen our conclusion, we performed the second derivative test. We found the second derivative, f''(x) = 6 + √(2) / (2x^(3/2)), and observed that it is positive for all x in the domain [0, ∞). This confirms that the function is concave up throughout its domain, and the critical point x = (√(2) / 6)^(2/3) corresponds to a global minimum.

Therefore, the least value of the function f(x) = 3x² - √(8x) + 32 is approximately 30.8345, and it occurs at x = (√(2) / 6)^(2/3). This problem highlights the power of calculus in solving optimization problems. By utilizing derivatives, we can systematically identify critical points and determine the minimum or maximum values of a function. The process involves careful algebraic manipulation, understanding the domain of the function, and applying the rules of differentiation. Furthermore, verification techniques, such as the second derivative test, provide additional confidence in our results. This problem also demonstrates the importance of a step-by-step approach to problem-solving. By breaking down the problem into smaller, manageable steps, we can effectively tackle complex mathematical challenges and arrive at accurate solutions.