Finding Local Maxima Of F(x) = -e^x(x+4) A Step-by-Step Guide
Introduction to Local Maxima
In the realm of calculus, understanding the behavior of functions is paramount. One critical aspect is identifying local maxima, which are points where a function reaches a peak within a specific interval. These points are crucial for optimization problems, curve sketching, and various applications in science and engineering. To find the local maximum or maxima of a function, we delve into the concepts of derivatives and critical points. This article focuses on a detailed exploration of how to determine the local maxima of the function $f(x) = -e^x(x+4)$. We will employ differential calculus techniques, emphasizing the first and second derivative tests, to rigorously identify the points at which the function attains its local maximum values. By understanding these methods, we gain valuable insights into the function’s behavior, enabling us to make informed decisions in related contexts. The process involves calculating the first derivative to find critical points, which are potential locations of maxima or minima. Then, we apply the second derivative test to ascertain the nature of these critical points, distinguishing between local maxima, local minima, and saddle points. Furthermore, we will discuss the theoretical underpinnings of these tests, ensuring a comprehensive understanding of why they work. The exploration will also extend to graphical representations, aiding in the visualization of the function's behavior and the identification of local maxima. This blend of analytical and graphical approaches offers a robust understanding of how to find and interpret local maxima.
Understanding the Function: f(x) = -e^x(x+4)
The function we are analyzing is $f(x) = -e^x(x+4)$. Before diving into the calculus, it’s essential to understand the nature of this function. This function is a product of two primary components: the exponential function $-e^x$ and the linear term $(x+4)$. The exponential term plays a significant role in dictating the function's behavior as $x$ varies. The negative sign in front of the exponential term means that the function will decay as $x$ increases, but the linear term $x+4$ introduces a twist, leading to a more complex interaction. To fully appreciate the function, consider its asymptotic behavior. As $x$ approaches negative infinity, $e^x$ approaches zero, causing the entire function to approach zero. However, as $x$ becomes very large, the exponential decay dominates, and the function approaches zero from the negative side. To find the local maximum/maxima of the function, we need to identify the critical points where the derivative equals zero or is undefined. These points are potential locations where the function changes direction, from increasing to decreasing, indicating a local maximum. The product of $-e^x$ and $(x+4)$ implies that we will need to use the product rule in differentiation. Graphing the function can also provide a visual representation of its behavior, showing peaks and valleys that correspond to local maxima and minima. In summary, the interplay between exponential decay and linear growth in $f(x) = -e^x(x+4)$ creates a function with a specific shape, making the identification of local maxima an interesting and mathematically rich problem.
Calculating the First Derivative
The first step in finding the local maxima of $f(x) = -e^x(x+4)$ is to compute its first derivative, $f'(x)$. The first derivative will help us identify the critical points of the function. To find $f'(x)$, we need to apply the product rule, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. In our case, let $u(x) = -e^x$ and $v(x) = (x+4)$. Then, the derivatives are $u'(x) = -e^x$ and $v'(x) = 1$. Applying the product rule, we get:
Simplifying this expression, we have:
We can factor out $-e^x$ from both terms:
Thus, the first derivative of the function $f(x) = -e^x(x+4)$ is $f'(x) = -e^x(x+5)$. The first derivative is a crucial tool because it tells us about the slope of the tangent line to the function at any given point. Critical points occur where the derivative is either zero or undefined. Since $e^x$ is never zero, the critical points of the function will be where the term $(x+5)$ equals zero. The derivative provides insights into where the function is increasing or decreasing, which are essential in locating local maxima and minima. In the next step, we will set the derivative equal to zero to solve for the critical points and determine their nature using further analysis.
Finding Critical Points
Now that we have the first derivative, $f'(x) = -e^x(x+5)$, we need to find the critical points of the function. Critical points occur where the first derivative is either equal to zero or undefined. Since $e^x$ is always positive and never equals zero, the term $-e^x$ will not contribute to the zeroes of the derivative. Therefore, we only need to consider the factor $(x+5)$. Setting the first derivative equal to zero, we get:
Since $-e^x$ is never zero, we solve for:
This gives us:
So, the critical point of the function $f(x) = -e^x(x+4)$ is $x = -5$. This point is significant because it is a potential location for a local maximum or minimum. To determine whether it's a local maximum, local minimum, or neither, we can use the first derivative test or the second derivative test. The critical point $x = -5$ divides the real number line into two intervals: $(-\infty, -5)$ and $(-5, \infty)$. We will analyze the sign of the first derivative in these intervals to understand the behavior of the function. In the interval $(-\infty, -5)$, $x$ is less than -5, so $(x+5)$ is negative. Since $-e^x$ is always negative, the product $-e^x(x+5)$ is positive. This means the function is increasing on this interval. In the interval $(-5, \infty)$, $x$ is greater than -5, so $(x+5)$ is positive. The product $-e^x(x+5)$ is thus negative, meaning the function is decreasing on this interval. These observations suggest that $x = -5$ is indeed a local maximum. To confirm this and further analyze the nature of the critical point, we can also use the second derivative test, which we will discuss in the following sections. Understanding the critical points and the intervals where the function increases or decreases is crucial in sketching the graph and solving optimization problems.
Applying the Second Derivative Test
To confirm whether the critical point $x = -5$ corresponds to a local maximum, we can use the second derivative test. This test involves finding the second derivative of the function and evaluating it at the critical point. If the second derivative is negative at the critical point, then the function has a local maximum there. If it's positive, the function has a local minimum. If it's zero, the test is inconclusive. First, let's find the second derivative of $f(x) = -e^x(x+4)$. We know that the first derivative is $f'(x) = -e^x(x+5)$. To find the second derivative, $f''(x)$, we again apply the product rule, letting $u(x) = -e^x$ and $v(x) = (x+5)$. Then, $u'(x) = -e^x$ and $v'(x) = 1$. Applying the product rule, we get:
Simplifying this expression, we have:
Factoring out $-e^x$, we get:
Now, we evaluate the second derivative at the critical point $x = -5$:
Since $-e^{-5}$ is negative, the second derivative is negative at $x = -5$. According to the second derivative test, this confirms that there is a local maximum at $x = -5$. This test provides a clear analytical method to verify the nature of critical points, which complements the information we gathered from analyzing the first derivative. By understanding the concavity of the function at the critical point, we can definitively classify it as a local maximum or minimum. In the next section, we will find the actual local maximum value of the function by plugging $x = -5$ into the original function.
Determining the Local Maximum Value
Now that we have confirmed the existence of a local maximum at $x = -5$, the next step is to find the actual value of the function at this point. This will give us the $y$-coordinate of the local maximum. To determine the local maximum value, we substitute $x = -5$ into the original function $f(x) = -e^x(x+4)$:
Thus, the local maximum value of the function is $e^{-5}$. This value is approximately equal to $0.0067$. The local maximum point is therefore at $(-5, e^{-5})$. This calculation completes the process of finding the local maximum of the function. We started by finding the first derivative to identify critical points, then used the second derivative test to confirm the nature of the critical point as a local maximum, and finally, calculated the value of the function at that point. Understanding this process is crucial for various applications in calculus and related fields. For instance, in optimization problems, we often need to find the maximum or minimum value of a function, and the techniques we've used here are directly applicable. In curve sketching, locating local maxima and minima helps us accurately represent the shape of the function. In physics and engineering, finding maximum and minimum values can help us analyze the behavior of systems under different conditions. The value $e^{-5}$ represents the highest point the function reaches in its local neighborhood around $x = -5$, and it's a key characteristic of the function’s behavior. In summary, the local maximum value of $f(x) = -e^x(x+4)$ is $e^{-5}$, and the point at which it occurs is $(-5, e^{-5})$.
Conclusion
In this article, we have thoroughly explored the process of finding the local maximum of the function $f(x) = -e^x(x+4)$. We began by understanding the nature of the function, which is a product of an exponential term and a linear term. This understanding laid the groundwork for the application of differential calculus techniques. The first step involved computing the first derivative, $f'(x) = -e^x(x+5)$, using the product rule. The first derivative is essential because it allows us to identify critical points, which are potential locations of local maxima or minima. We found the critical point by setting the first derivative equal to zero and solving for $x$, resulting in $x = -5$. To determine the nature of this critical point, we employed the second derivative test. We calculated the second derivative, $f''(x) = -e^x(x+6)$, and evaluated it at $x = -5$. The negative value of the second derivative, $f''(-5) = -e^{-5}$, confirmed that $x = -5$ corresponds to a local maximum. Finally, we calculated the local maximum value by substituting $x = -5$ back into the original function, obtaining $f(-5) = e^{-5}$. Thus, the local maximum of the function occurs at the point $(-5, e^{-5})$. This process illustrates a systematic approach to finding local maxima, which is a fundamental concept in calculus. Understanding local maxima and minima is crucial for various applications, including optimization problems, curve sketching, and real-world applications in physics, engineering, and economics. The combination of analytical techniques, such as finding derivatives and applying tests, provides a robust method for analyzing the behavior of functions. By mastering these techniques, we can gain valuable insights into the properties and characteristics of mathematical functions, enabling us to solve complex problems and make informed decisions. The ability to find the local maximum/maxima of functions like $f(x) = -e^x(x+4)$ is a cornerstone of mathematical analysis and its practical applications.
