Finding The Minimum Value Of F(x) = 3x² - √(8x) + 32
Finding the least value of a function is a fundamental problem in calculus and optimization. This article delves into a detailed exploration of the function f(x) = 3x² - √(8x) + 32, providing a step-by-step approach to determine its minimum value. Understanding the behavior of functions, especially their extreme values, is crucial in various fields like engineering, economics, and computer science. In this guide, we'll cover the necessary calculus techniques, including differentiation and critical point analysis, to effectively solve this optimization problem.
1. Understanding the Function
Before diving into the calculations, it's essential to understand the nature of the function f(x) = 3x² - √(8x) + 32. This function combines a quadratic term (3x²) with a square root term (-√(8x)) and a constant (32). The quadratic term contributes to a parabolic shape, while the square root term introduces a non-linear element. The domain of the function is restricted by the square root, requiring x ≥ 0. This means we only need to consider non-negative values of x when looking for the least value. Understanding the domain is the first critical step in any optimization problem.
The quadratic component, 3x², increases as x moves away from zero in either direction, but since we are constrained to x ≥ 0, it only increases as x increases. The square root component, √(8x), also increases as x increases, but it is being subtracted from the quadratic term. This creates an interplay between the two terms, which is crucial in determining where the minimum value occurs. The constant term, +32, simply shifts the entire function upwards but does not affect the location of the minimum. Therefore, our primary focus will be on how the 3x² and √(8x) terms interact.
To visualize this interaction, imagine starting at x = 0. At this point, the quadratic term is zero, and the square root term is also zero, so the function value is simply 32. As x increases slightly, both the quadratic and square root terms start to increase. However, initially, the square root term might increase more rapidly, causing the function value to decrease. As x continues to increase, the quadratic term will eventually dominate, causing the function value to increase. This suggests that there is a point where the decrease due to the square root term is balanced by the increase due to the quadratic term, leading to a minimum value. This intuition is precisely what we will formalize using calculus.
Furthermore, let's consider the behavior of the function as x approaches infinity. The quadratic term 3x² will grow much faster than the square root term √(8x). This means that as x becomes very large, the function will be dominated by the quadratic term and will increase without bound. This observation confirms that the function does not have a maximum value. Therefore, our focus is indeed on finding the minimum value, which we expect to occur at some finite value of x.
In summary, by understanding the individual components of the function and their interactions, we have developed a strong intuition for the behavior of f(x). We know that the function is defined for x ≥ 0, it starts at 32 when x = 0, and it likely has a minimum value at some positive value of x. The next step is to use calculus to find this minimum value precisely.
2. Finding the Derivative
The core of finding the minimum value of a function lies in calculus, specifically using derivatives. The derivative of a function gives us the rate of change of the function, and the points where the derivative is zero or undefined are critical points, which may correspond to local minima, local maxima, or saddle points. The first step in our quest to find the least value of f(x) = 3x² - √(8x) + 32 is to compute its derivative, f'(x). This will help us identify the critical points of the function.
To find the derivative, we apply the power rule and chain rule of differentiation. Recall that the power rule states that the derivative of xⁿ is nxⁿ⁻¹, and the chain rule is used when differentiating composite functions. First, let's rewrite the function to make differentiation easier:
f(x) = 3x² - (8x)^(1/2) + 32
Now, we can differentiate each term separately. The derivative of 3x² is 6x. The derivative of (8x)^(1/2) can be found using the chain rule: the derivative of u^(1/2) is (1/2)u^(-1/2), and the derivative of 8x is 8. Therefore, the derivative of (8x)^(1/2) is (1/2)(8x)^(-1/2) * 8 = 4(8x)^(-1/2). The derivative of the constant term 32 is zero.
Combining these results, we get the derivative of f(x):
f'(x) = 6x - 4(8x)^(-1/2)
We can simplify this expression further:
f'(x) = 6x - 4 / √(8x)
To make it even more manageable, let's rewrite the square root term:
√(8x) = √(8) * √x = 2√2 * √x
So, the derivative becomes:
f'(x) = 6x - 4 / (2√2 * √x)
f'(x) = 6x - √2 / √x
This expression for f'(x) is crucial. It represents the slope of the tangent line to the graph of f(x) at any point x. The points where f'(x) = 0 are the critical points of the function. These points are potential locations for local minima, local maxima, or saddle points. The next step is to find these critical points by setting f'(x) equal to zero and solving for x.
The process of finding the derivative is a fundamental step in optimization problems. A correct derivative is essential for identifying critical points and subsequently determining the minimum or maximum values of the function. In our case, we have obtained a derivative that involves both a linear term (6x) and a term with a negative fractional exponent (-√2 / √x). This combination suggests that there will be a balancing point where the positive linear term counteracts the negative fractional exponent term, leading to a critical point. This intuition guides us as we proceed to find the critical points.
3. Finding Critical Points
Now that we have the derivative, f'(x) = 6x - √2 / √x, the next step is to find the critical points of the function. Critical points occur where the derivative is either equal to zero or undefined. These points are crucial because they are potential locations for local minima, local maxima, or saddle points. To find the critical points, we set f'(x) = 0 and solve for x.
Setting the derivative equal to zero gives us:
6x - √2 / √x = 0
To solve this equation, we can start by adding √2 / √x to both sides:
6x = √2 / √x
Next, multiply both sides by √x to eliminate the fraction:
6x√x = √2
We can rewrite x√x as x^(3/2), so the equation becomes:
6x^(3/2) = √2
Now, 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 is the value of x that makes the derivative zero. Now, let's simplify this expression. We can rewrite √2 as 2^(1/2), so:
x = (2^(1/2) / 6)^(2/3)
To simplify further, we can rewrite 6 as 2 * 3, so:
x = (2^(1/2) / (2 * 3))^(2/3)
x = (2^(-1/2) / 3)^(2/3)
Applying the exponent to both terms in the fraction, we get:
x = 2^(-1/3) / 3^(2/3)
This value of x is a critical point of the function. We also need to consider 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.
In summary, we have found two critical points: x = 0 and x = 2^(-1/3) / 3^(2/3). These points are the key to finding the minimum value of the function. The next step is to determine whether these critical points correspond to a minimum, a maximum, or a saddle point. We will use the second derivative test to make this determination.
4. Determining the Nature of Critical Points Using the Second Derivative Test
Having identified the critical points, x = 0 and x = 2^(-1/3) / 3^(2/3), we now need to determine the nature of these points. Are they local minima, local maxima, or saddle points? One effective method for this is the second derivative test. The second derivative test involves finding the second derivative of the function, f''(x), and evaluating it at the critical points. The sign of f''(x) at a critical point tells us about the concavity of the function at that point.
If f''(x) > 0, the function is concave up, and the critical point is a local minimum. If f''(x) < 0, the function is concave down, and the critical point is a local maximum. If f''(x) = 0, the test is inconclusive, and we need to use other methods to determine the nature of the critical point.
Let's find the second derivative of f(x). We start with the first derivative:
f'(x) = 6x - √2 / √x = 6x - √2 * x^(-1/2)
Now, we differentiate f'(x) to find f''(x). The derivative of 6x is 6. The derivative of -√2 * x^(-1/2) can be found using the power rule: (-1/2) * (-√2) * x^(-3/2) = (√2 / 2) * x^(-3/2). So, the second derivative is:
f''(x) = 6 + (√2 / 2) * x^(-3/2)
f''(x) = 6 + √2 / (2x^(3/2))
Now, we evaluate f''(x) at the critical points. First, let's consider x = 0. The second derivative f''(x) = 6 + √2 / (2x^(3/2)) is undefined at x = 0 because we would be dividing by zero. Therefore, the second derivative test is inconclusive at x = 0. We will need to use another method to analyze this critical point later.
Next, we evaluate f''(x) at the other critical point, x = 2^(-1/3) / 3^(2/3). Let's call this value x₀:
x₀ = 2^(-1/3) / 3^(2/3)
We need to determine the sign of f''(x₀):
f''(x₀) = 6 + √2 / (2(2^(-1/3) / 3(2/3))(3/2))
f''(x₀) = 6 + √2 / (2 * (2(-1/3))(3/2) / (3(2/3))(3/2))
f''(x₀) = 6 + √2 / (2 * 2^(-1/2) / 3)
f''(x₀) = 6 + √2 / (2 / (√2 * 3))
f''(x₀) = 6 + √2 * (√2 * 3) / 2
f''(x₀) = 6 + 2 * 3 / 2
f''(x₀) = 6 + 3
f''(x₀) = 9
Since f''(x₀) = 9 > 0, the function is concave up at x = x₀, which means that x₀ = 2^(-1/3) / 3^(2/3) is a local minimum. This is a significant finding. We have identified a critical point where the function has a minimum value.
5. Evaluating the Function at the Critical Points and Boundaries
Having determined that x = 2^(-1/3) / 3^(2/3) is a local minimum using the second derivative test, we now need to find the least value of the function. This involves evaluating the function at the critical points and considering the behavior of the function at the boundaries of its domain. Recall that the domain of f(x) = 3x² - √(8x) + 32 is x ≥ 0.
We have two critical points to consider: x = 0 and x = x₀ = 2^(-1/3) / 3^(2/3). First, let's evaluate the function at x = 0:
f(0) = 3(0)² - √(8 * 0) + 32 = 32
Next, we need to evaluate the function at x = x₀. This calculation is a bit more involved:
f(x₀) = 3(2^(-1/3) / 3^(2/3))² - √(8 * (2^(-1/3) / 3^(2/3))) + 32
Let's simplify this expression step by step. First, square the term inside the first parenthesis:
(2^(-1/3) / 3^(2/3))² = 2^(-2/3) / 3^(4/3)
So, the first term becomes:
3 * (2^(-2/3) / 3^(4/3)) = 2^(-2/3) / 3^(1/3)
Now, let's simplify the term inside the square root:
8 * (2^(-1/3) / 3^(2/3)) = 2³ * 2^(-1/3) / 3^(2/3) = 2^(8/3) / 3^(2/3)
Taking the square root:
√(2^(8/3) / 3^(2/3)) = 2^(4/3) / 3^(1/3)
Since √(8x) = √(8) * √x = 2√2 * √x, we can rewrite the second term as:
√(8 * (2^(-1/3) / 3^(2/3))) = √(8) * √(2^(-1/3) / 3^(2/3))
√(8 * (2^(-1/3) / 3^(2/3))) = 2√2 * 2^(-1/6) / 3^(1/3)
√(8 * (2^(-1/3) / 3^(2/3))) = 2^(1 + 1/2 - 1/6) / 3^(1/3) = 2^(4/3) / 3^(1/3)
Now, we can substitute these simplified terms back into the expression for f(x₀):
f(x₀) = 2^(-2/3) / 3^(1/3) - 2^(4/3) / 3^(1/3) + 32
To combine the first two terms, we need a common denominator. We can multiply the first term by 2²/2:
2^(-2/3) / 3^(1/3) = 1/ (2^(2/3) * 3^(1/3))
And rewrite the second term similarly:
2^(4/3) / 3^(1/3) = 2² / (2^(-2/3) * 3^(1/3))
Then we can see:
f(x₀) = 3 * 2^(-2/3) / 3^(2/3) - √(2^(8/3) / 3^(2/3)) + 32
Let's simplify x₀ = (√2 / 6)^(2/3).
x₀^(3/2) = √2 / 6 = √2 / (2 * 3), multiply by √2/√2, x₀^(3/2) = 2 / (2 * 3 * √2) = 1 / (3√2)
f'(x) = 6x - √2 / √x = (6x^(3/2) - √2) / √x
f'(x₀) = 6 * √2 / 6 - √2 / √((√2 / 6)^(2/3)) = 0
Then,
6 x₀^(3/2) = √2
x₀^(3/2) = √2 / 6
f(x₀) = 3x₀² - √(8x₀) + 32
We can derive from first derivative result equation,
6 x₀^(3/2) = √2
Square both side,
36 x₀³ = 2
x₀³ = 1 / 18
x₀ = 1 / (18^(1/3))
And,
3x₀² = 3 * (2^(-2/3) / 3^(4/3)) = 1 / (2^(2/3) * 3^(1/3)) = 1 / (4^(1/3) * 3^(1/3)) = 1 / 12^(1/3)
Also,
√(8x₀) = √(8 / 18^(1/3)) = 2√2 / (18^(1/6))
√(8x₀) = √(8) x₀^(1/2) = 2√2 * (1 / 18(1/3))(1/2) = 2√2 / (18^(1/6))
√(8x₀) = 2^(3/2) / (2 * 32)(1/6) = 2^(7/6) / 3^(1/3)
Then,
f(x₀) = 1 / 12^(1/3) - 2^(7/6) / 3^(1/3) + 32 ≈ 29.4025633715
Comparing f(0) = 32 and f(x₀) ≈ 29.402, we see that f(x₀) is smaller. Therefore, the least value of the function occurs at x = x₀ = 2^(-1/3) / 3^(2/3), and the least value is approximately 29.402.
6. Conclusion: The Least Value of the Function
In conclusion, we have successfully found the least value of the function f(x) = 3x² - √(8x) + 32. We achieved this by systematically applying the principles of calculus and optimization. Here's a recap of the steps we took:
- Understanding the Function: We analyzed the function's components and determined its domain, x ≥ 0. We also developed an intuition for its behavior, noting the interplay between the quadratic and square root terms.
- Finding the Derivative: We computed the first derivative of the function, f'(x) = 6x - √2 / √x, which represents the rate of change of the function.
- Finding Critical Points: We identified the critical points by setting f'(x) = 0 and solving for x. We found two critical points: x = 0 and x = 2^(-1/3) / 3^(2/3).
- Determining the Nature of Critical Points: We used the second derivative test to determine the nature of the critical points. We found that x = 2^(-1/3) / 3^(2/3) is a local minimum.
- Evaluating the Function: We evaluated the function at the critical points and found that the least value occurs at x = 2^(-1/3) / 3^(2/3), with a function value of approximately 29.402.
Therefore, the least value of the function f(x) = 3x² - √(8x) + 32 is approximately 29.402, which occurs at x = 2^(-1/3) / 3^(2/3). This comprehensive approach demonstrates the power of calculus in solving optimization problems and provides a clear methodology for finding the minimum values of functions.
This problem illustrates a typical optimization scenario where understanding the function's behavior, finding critical points, and using the second derivative test are essential steps. The final evaluation at critical points and boundaries ensures that we identify the absolute minimum value. This methodology can be applied to a wide range of optimization problems in various fields, making it a valuable skill for students and professionals alike.
