Finding The Absolute Extremum Of F(x) = -xe^(-x/2) A Calculus Exploration

In the realm of mathematical analysis, identifying extreme values of functions is a cornerstone concept with far-reaching applications. From optimizing engineering designs to predicting economic trends, the ability to pinpoint maxima and minima is invaluable. This article delves into the function f(x) = -xe^(-x/2), a seemingly simple expression that reveals intricate behavior upon closer inspection. Our primary objective is to determine whether this function possesses an absolute extremum, and if so, to precisely locate its position. By employing the tools of calculus, such as differentiation and the analysis of critical points, we embark on a journey to unravel the function's characteristics and unveil its extreme values. This exploration not only enhances our understanding of this specific function but also reinforces the fundamental principles of extremum determination, applicable across a diverse spectrum of mathematical and scientific disciplines. Understanding the behavior of functions like f(x) = -xe^(-x/2) is crucial in various fields, including physics, engineering, and economics, where optimization problems frequently arise. For example, in physics, this function could model the decay of a signal over time, while in economics, it might represent the profit margin of a product as a function of investment. Therefore, a thorough analysis of its extreme values is not merely an academic exercise but a practical necessity. The journey we undertake will illuminate the interplay between algebraic expressions and their graphical representations, fostering a deeper appreciation for the elegance and power of mathematical analysis. So, let's embark on this exploration, armed with the principles of calculus and a thirst for mathematical discovery.

Unveiling the Critical Points

To locate the absolute extremum of the function f(x) = -xe^(-x/2), our first crucial step involves identifying its critical points. These points, where the function's derivative equals zero or is undefined, hold the key to understanding the function's behavior, particularly its local maxima and minima. The derivative, a fundamental concept in calculus, provides us with the instantaneous rate of change of the function at any given point. By setting the derivative to zero, we effectively pinpoint the locations where the function's slope is horizontal, indicating potential turning points. To embark on this process, we must first compute the derivative of f(x). Applying the product rule, a cornerstone of differentiation, we carefully dissect the function into its constituent parts: -x and e^(-x/2). The product rule, which states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second, guides our hand as we navigate the intricacies of differentiation. With the product rule as our compass, we proceed to differentiate each part. The derivative of -x is simply -1, a straightforward application of the power rule. However, the derivative of e^(-x/2) requires a touch of the chain rule, another essential tool in our calculus arsenal. The chain rule, designed for composite functions, allows us to peel away the layers of the function, differentiating the outer layer first and then working our way inward. Applying the chain rule, we find that the derivative of e^(-x/2) is (-1/2)e^(-x/2). Now, with the derivatives of each part in hand, we can assemble the complete derivative of f(x) using the product rule. This process yields f'(x) = -e^(-x/2) + (x/2)e^(-x/2). This expression, the fruit of our differentiation labors, is the key to unlocking the function's critical points. To find these critical points, we set f'(x) equal to zero and embark on the task of solving for x. This equation, -e^(-x/2) + (x/2)e^(-x/2) = 0, may appear daunting at first glance, but with careful algebraic manipulation, we can tame its complexity. Factoring out the common factor of e^(-x/2), we simplify the equation to e^(-x/2)(-1 + x/2) = 0. This factorization reveals a crucial insight: the product of two terms equals zero if and only if at least one of the terms is zero. This principle, a cornerstone of algebraic problem-solving, allows us to break the equation into two simpler cases. The first case, e^(-x/2) = 0, is quickly dismissed since the exponential function is never zero for any real value of x. This leaves us with the second case: -1 + x/2 = 0. Solving this simple linear equation, we arrive at the critical point x = 2. This solitary critical point, a beacon in our mathematical landscape, marks a potential location for an extremum. However, to definitively determine whether this critical point corresponds to a maximum, a minimum, or neither, we must delve deeper into the function's behavior.

Ascertaining the Nature of the Extremum

Having pinpointed the critical point at x = 2 for the function f(x) = -xe^(-x/2), our next imperative is to discern the nature of this point. Does it represent a local maximum, a local minimum, or perhaps an inflection point? To answer this question, we turn to the powerful tools of calculus, specifically the second derivative test. The second derivative, a measure of the rate of change of the slope of the function, provides us with invaluable information about the function's concavity. A positive second derivative indicates that the function is concave up, resembling a smile, while a negative second derivative signifies concavity down, akin to a frown. At a local minimum, the function's concavity is upward, while at a local maximum, the concavity is downward. Thus, the sign of the second derivative at a critical point provides a direct indication of the nature of the extremum. To embark on this investigation, we must first compute the second derivative of f(x). Recalling that the first derivative is f'(x) = -e^(-x/2) + (x/2)e^(-x/2), we differentiate this expression once more, employing the product rule and chain rule as needed. The derivative of -e^(-x/2) is (1/2)e^(-x/2), while the derivative of (x/2)e^(-x/2) requires the product rule. Applying the product rule, we find the derivative of (x/2)e^(-x/2) to be (1/2)e^(-x/2) - (x/4)e^(-x/2). Combining these results, we obtain the second derivative: f''(x) = (1/2)e^(-x/2) + (1/2)e^(-x/2) - (x/4)e^(-x/2), which simplifies to f''(x) = e^(-x/2) - (x/4)e^(-x/2). Now, with the second derivative in hand, we evaluate it at the critical point x = 2. This evaluation yields f''(2) = e^(-2/2) - (2/4)e^(-2/2) = e^(-1) - (1/2)e^(-1) = (1/2)e^(-1). Since e^(-1) is a positive value, it follows that f''(2) is also positive. This positive second derivative at x = 2 decisively indicates that the function is concave up at this point, confirming that we have indeed located a local minimum. However, our quest extends beyond merely identifying a local extremum; we seek the absolute extremum, the function's global minimum or maximum value. To ascertain whether the local minimum at x = 2 is also the absolute minimum, we must consider the function's behavior across its entire domain. As x approaches positive infinity, the exponential term e^(-x/2) decays to zero more rapidly than the linear term -x grows, causing the function f(x) to approach zero. Conversely, as x approaches negative infinity, the exponential term dominates, driving the function towards positive infinity. This analysis reveals that the function has no absolute maximum. However, since the function approaches zero as x approaches infinity and tends towards positive infinity as x approaches negative infinity, the local minimum at x = 2 is indeed the absolute minimum. To determine the absolute minimum value, we simply evaluate the function at x = 2: f(2) = -2e^(-2/2) = -2e^(-1) = -2/e. Thus, we have definitively established that the function f(x) = -xe^(-x/2) possesses an absolute minimum of -2/e at x = 2. This journey, guided by the principles of calculus, has unveiled the extremum of this function, highlighting the power of mathematical analysis in understanding the behavior of complex expressions.

Determining the Absolute Extremum

Having established that the function f(x) = -xe^(-x/2) has a local minimum at x = 2, our final task is to confirm whether this local minimum is also the absolute extremum. This requires a comprehensive understanding of the function's behavior across its entire domain, not just in the vicinity of the critical point. To achieve this, we delve into the concept of limits, a cornerstone of calculus that allows us to analyze the function's behavior as its input approaches infinity or negative infinity. By examining these limiting cases, we can gain insights into the function's overall trend and determine whether the local minimum is indeed the lowest point on the function's graph. First, let us consider the limit as x approaches positive infinity. As x grows without bound, the term -x in the function f(x) = -xe^(-x/2) becomes increasingly negative. However, the exponential term e^(-x/2) decays towards zero at a much faster rate. This interplay between the linear term and the exponential term is crucial. The exponential decay dominates the linear growth, effectively