Finding The Minimum Value Of F(x) = Max(11x, 60 - 2x²) And Doubling It

Jul 14, 2025 by Jeany 71 views

In this article, we delve into the fascinating world of mathematical functions, specifically focusing on finding the minimum value of a function defined as the maximum of two other functions. The function in question is f(x) = max(11x, 60 - 2x²), where x is any positive real number. Our goal is to determine the minimum possible value of f(x) and then double it. This exploration will involve understanding the behavior of both linear and quadratic functions, finding their points of intersection, and applying the concept of maximum value within a specific domain. Let's embark on this mathematical journey!

Understanding the Functions

To effectively tackle this problem, we need to understand the individual functions that make up f(x). We have two functions:

g(x) = 11x: This is a linear function with a positive slope of 11. This means that as x increases, g(x) also increases linearly. The graph of this function is a straight line passing through the origin with a steep upward slope. Understanding linear functions is crucial as they represent direct relationships and are fundamental in various mathematical and real-world applications.
h(x) = 60 - 2x²: This is a quadratic function representing a parabola that opens downwards. The coefficient of the x² term is negative (-2), indicating the downward concavity. The vertex of the parabola represents the maximum point of this function. As x moves away from the vertex in either direction, the value of h(x) decreases. Quadratic functions are essential in modeling parabolic trajectories, projectile motion, and various optimization problems.

The function f(x) is defined as the maximum of these two functions at any given value of x. This means that for each x, f(x) will take the higher value between g(x) and h(x). This “maximum” definition introduces a piecewise nature to f(x), making it crucial to identify the intervals where each of g(x) and h(x) dominates. To fully grasp this, we must explore where these two functions intersect and how their values compare on either side of these intersection points. Understanding the interplay between these functions is the key to unraveling the minimum value of f(x).

Finding the Intersection Points

The intersection points of the two functions, g(x) = 11x and h(x) = 60 - 2x², are crucial for determining the intervals where each function is greater. To find these points, we need to solve the equation g(x) = h(x), which means setting the two functions equal to each other:

11x = 60 - 2x²

Rearranging the equation, we get a quadratic equation:

2x² + 11x - 60 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Where a = 2, b = 11, and c = -60. Plugging in these values, we get:

x = (-11 ± √(11² - 4 * 2 * -60)) / (2 * 2) x = (-11 ± √(121 + 480)) / 4 x = (-11 ± √601) / 4

This gives us two solutions:

x₁ = (-11 + √601) / 4 ≈ (-11 + 24.515) / 4 ≈ 3.379 x₂ = (-11 - √601) / 4 ≈ (-11 - 24.515) / 4 ≈ -8.879

Since we are only considering positive real numbers for x, we can disregard the negative solution. Therefore, the intersection point occurs at approximately x ≈ 3.379. At this point, the values of g(x) and h(x) are equal. This intersection point is a critical threshold. For x values less than 3.379, one function will be greater, and for x values greater than 3.379, the other function will dominate. Accurately determining this intersection is paramount to understanding the behavior of f(x) and ultimately finding its minimum value. This highlights the significance of solving quadratic equations and interpreting their solutions in the context of the problem.

Analyzing the Intervals

Now that we've found the intersection point at approximately x ≈ 3.379, we can analyze the behavior of the functions g(x) = 11x and h(x) = 60 - 2x² in the intervals defined by this point. This analysis will help us understand which function contributes to the maximum value in f(x) = max(11x, 60 - 2x²) for different ranges of x.

For 0 < x < 3.379: In this interval, the quadratic function h(x) = 60 - 2x² is greater than the linear function g(x) = 11x. To see this, consider a value less than 3.379, say x = 2. At this point, g(2) = 11 * 2 = 22, and h(2) = 60 - 2 * 2² = 60 - 8 = 52. So, h(x) is larger. This means that f(x) = h(x) = 60 - 2x² in this interval. The f(x) will follow the curve of the downward-opening parabola.
For x > 3.379: In this interval, the linear function g(x) = 11x is greater than the quadratic function h(x) = 60 - 2x². For instance, at x = 4, g(4) = 11 * 4 = 44, and h(4) = 60 - 2 * 4² = 60 - 32 = 28. Thus, g(x) dominates. Consequently, f(x) = g(x) = 11x in this interval. f(x) here will be a straight line with a positive slope.

At the intersection point x ≈ 3.379, both functions have the same value, so f(x) transitions from following the quadratic curve to the linear line. The minimum value of f(x) will occur at this transition point, as the quadratic function is decreasing up to this point, and the linear function is increasing from this point onwards. This interval analysis is a crucial step in identifying the behavior of f(x) and pinpointing where the minimum value lies. By understanding which function dictates the value of f(x) in each interval, we can effectively determine the overall minimum.

Finding the Minimum Value of f(x)

The minimum value of f(x) = max(11x, 60 - 2x²) will occur at the intersection point of the two functions g(x) = 11x and h(x) = 60 - 2x². We've already found this intersection point to be approximately x ≈ 3.379. To find the minimum value of f(x), we can plug this x value into either g(x) or h(x) since they are equal at this point. Let's use g(x):

f(3.379) = 11 * 3.379 ≈ 37.169

Therefore, the minimum value of f(x) is approximately 37.169. This minimum value represents the lowest point on the piecewise function f(x), where the transition occurs from the quadratic function dominating to the linear function taking over. To visualize this, imagine the graph of f(x): it initially follows the curve of the parabola h(x), decreasing until it reaches the intersection point, and then it follows the straight line of g(x), increasing thereafter. The minimum point is where these two segments meet.

Doubling the Minimum Value

The final step is to double the minimum value of f(x) that we found. We determined that the minimum value is approximately 37.169. So, doubling this value:

2 * 37.169 ≈ 74.338

Therefore, the double of the minimum possible value of f(x) is approximately 74.338. This result is significant because it answers the original question posed. By understanding the individual functions, finding their intersection, and analyzing their behavior in different intervals, we successfully determined the minimum value of f(x) and its double. This process showcases the power of mathematical analysis in solving complex problems involving piecewise functions and optimization.

Conclusion

In summary, we have successfully found the double of the minimum possible value of the function f(x) = max(11x, 60 - 2x²). This involved a detailed analysis of the linear function g(x) = 11x and the quadratic function h(x) = 60 - 2x², finding their intersection point, and understanding their behavior in different intervals. We determined that the minimum value of f(x) occurs at the intersection point, which is approximately 37.169. Doubling this value, we arrived at the final answer of approximately 74.338.

This exploration has highlighted the importance of understanding different types of functions, solving equations, and applying mathematical concepts to real-world problems. The process of finding the minimum value of a function defined as the maximum of two other functions is a powerful technique that can be applied in various fields, including optimization, engineering, and economics. By mastering these fundamental mathematical skills, we can confidently tackle complex problems and gain valuable insights into the world around us.