Finding Critical Points And Classifying With The Second Derivative Test For F(x) = -e^x(x+4)
Introduction
In this article, we will delve into the process of locating the critical points of the function f(x) = -e^x(x+4). Understanding critical points is crucial in calculus as they provide valuable insights into the behavior of a function. Specifically, we'll identify where the function's rate of change is zero or undefined, which can indicate potential local maxima, local minima, or saddle points. Furthermore, we will employ the Second Derivative Test to classify these critical points, determining whether they correspond to local maxima, local minima, or neither. This comprehensive analysis will not only enhance our understanding of the function's graph but also demonstrate the practical application of calculus principles in function analysis.
Determining the Critical Points
To find the critical points of the function f(x) = -e^x(x+4), we first need to compute its first derivative, f'(x). This involves applying the product rule and chain rule of differentiation. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. The chain rule is used when differentiating a composite function. In this case, we have:
f(x) = -e^x(x+4)
Applying the product rule, we get:
f'(x) = -[e^x(x+4) + e^x(1)]
Simplifying this expression, we have:
f'(x) = -e^x(x+4) - e^x
f'(x) = -e^x(x+4+1)
f'(x) = -e^x(x+5)
Critical points occur where the first derivative f'(x) is either equal to zero or undefined. The exponential function e^x is never zero, but it is defined for all real numbers. Therefore, the critical points occur when:
-e^x(x+5) = 0
Since -e^x is never zero, we only need to solve:
x+5 = 0
This gives us the critical point:
x = -5
Thus, the critical point of the function f(x) = -e^x(x+4) is x = -5. This point is crucial because it marks a potential change in the function's behavior, such as a transition from increasing to decreasing or vice versa. To fully understand the nature of this critical point, we need to determine whether it corresponds to a local maximum, a local minimum, or neither, which is where the Second Derivative Test comes into play.
Applying the Second Derivative Test
To determine whether the critical point at x = -5 corresponds to a local maximum, local minimum, or neither, we use the Second Derivative Test. This test involves computing the second derivative of the function, f''(x), and evaluating it at the critical point. The sign of f''(-5) will tell us about the concavity of the function at x = -5, which in turn indicates the nature of the critical point.
First, we need to find the second derivative, f''(x). We start with the first derivative we found earlier:
f'(x) = -e^x(x+5)
Applying the product rule again, we get:
f''(x) = -[e^x(x+5) + e^x(1)]
Simplifying this expression, we have:
f''(x) = -e^x(x+5) - e^x
f''(x) = -e^x(x+5+1)
f''(x) = -e^x(x+6)
Now, we evaluate f''(x) at the critical point x = -5:
f''(-5) = -e^{-5}(-5+6)
f''(-5) = -e^{-5}(1)
f''(-5) = -e^{-5}
Since e^{-5} is a positive number, -e^{-5} is negative. According to the Second Derivative Test:
- If f''(c) > 0, then f(x) has a local minimum at x = c.
- If f''(c) < 0, then f(x) has a local maximum at x = c.
- If f''(c) = 0 or f''(c) is undefined, the test is inconclusive.
In our case, f''(-5) = -e^{-5} < 0, which means that the function f(x) has a local maximum at x = -5. This indicates that the function reaches a peak at this point, where it changes from increasing to decreasing. This conclusion is a direct application of the Second Derivative Test, which is a powerful tool for analyzing the behavior of functions around their critical points.
Conclusion
In summary, we have successfully located the critical point of the function f(x) = -e^x(x+4) and used the Second Derivative Test to determine its nature. By calculating the first derivative, f'(x) = -e^x(x+5), we found that the critical point occurs at x = -5. This is the point where the function's rate of change is zero, making it a potential location for a local extremum.
Next, we computed the second derivative, f''(x) = -e^x(x+6), and evaluated it at the critical point. We found that f''(-5) = -e^{-5}, which is a negative value. According to the Second Derivative Test, a negative second derivative at a critical point indicates that the function has a local maximum at that point. Therefore, we concluded that the function f(x) = -e^x(x+4) has a local maximum at x = -5.
This analysis demonstrates the importance of critical points and the Second Derivative Test in understanding the behavior of functions. By locating critical points and classifying them using the Second Derivative Test, we can gain valuable insights into the function's graph, including where it reaches local maxima and minima. This knowledge is essential in various applications of calculus, such as optimization problems, curve sketching, and the analysis of rates of change. Understanding these concepts allows us to make informed predictions about the function's behavior and its applications in real-world scenarios.
