Extreme Value Theorem Application To Y=ln(x+2)/(x-2) Intervals

The Extreme Value Theorem is a cornerstone concept in calculus, providing crucial insights into the behavior of continuous functions on closed intervals. This article delves into the application of the Extreme Value Theorem to the function $y = \frac{\ln(x+2)}{x-2}$, examining the intervals [-3, -1], [-1, 1], and [1, 3] to determine where the theorem's conditions are met. Understanding the nuances of continuity and closed intervals is paramount to correctly applying this theorem. In this comprehensive analysis, we will meticulously dissect each interval, providing a clear and accessible explanation suitable for students and enthusiasts alike.

Understanding the Extreme Value Theorem

At its core, the Extreme Value Theorem states that if a function is continuous on a closed interval [a, b], then the function must attain both a maximum and a minimum value on that interval. This seemingly simple statement has profound implications in optimization problems and the study of function behavior. To fully grasp the theorem, it's essential to understand its two key components: continuity and closed intervals. A function is said to be continuous on an interval if its graph can be drawn without lifting the pen, meaning there are no breaks, jumps, or vertical asymptotes within the interval. A closed interval [a, b] includes both endpoints, a and b, which is critical for the theorem's validity. If either continuity or the closed interval condition is not met, the Extreme Value Theorem cannot be applied.

The significance of the Extreme Value Theorem lies in its ability to guarantee the existence of extreme values. Without this guarantee, finding the maximum and minimum values of a function becomes a much more challenging task. For instance, consider a function with a discontinuity within the interval of interest. The function might approach a certain value but never actually attain it, thus lacking a true maximum or minimum. Similarly, on an open interval, a function could increase or decrease indefinitely as it approaches the endpoint, again without ever reaching a maximum or minimum. The theorem provides a powerful tool for ensuring that such scenarios do not occur when dealing with continuous functions on closed intervals. The implications extend to various fields, including physics, engineering, and economics, where optimization problems are frequently encountered.

Analyzing the Function y = ln(x+2)/(x-2)

The function under consideration is $y = \frac{\ln(x+2)}{x-2}$. To apply the Extreme Value Theorem, we must first ascertain the function's continuity. The function has two potential sources of discontinuity: the natural logarithm and the denominator. The natural logarithm, $\ln(x+2)$, is only defined for $x+2 > 0$, which means $x > -2$. The denominator, $x-2$, introduces a vertical asymptote at $x = 2$. Therefore, the function is continuous on its domain, which is $(-2, 2) \cup (2, \infty)$. This initial assessment of the function's domain and continuity is crucial for determining the applicability of the Extreme Value Theorem on the given intervals.

Delving deeper into the function's characteristics, the logarithmic component, $\ln(x+2)$, dictates that the function is only defined for inputs greater than -2. This immediately rules out any interval that includes or extends beyond this boundary. The denominator, $x-2$, introduces a critical point at $x=2$, where the function becomes undefined due to division by zero. This point acts as a vertical asymptote, further restricting the intervals on which the Extreme Value Theorem can be applied. Understanding these restrictions is paramount. We must examine each provided interval in the context of these restrictions to ensure the function's continuity and the interval's closed nature. The interplay between the logarithmic domain restriction and the vertical asymptote at $x=2$ will ultimately dictate whether the Extreme Value Theorem can be successfully applied.

Interval I: [-3, -1]

For interval I, [-3, -1], we must check if the function $y = \frac{\ln(x+2)}{x-2}$ is continuous. Recall that the natural logarithm, $\ln(x+2)$, is only defined for $x > -2$. Since the interval [-3, -1] includes values less than -2 (specifically, values between -3 and -2), the natural logarithm is not defined for the entire interval. Consequently, the function is not continuous on [-3, -1]. This lack of continuity means the Extreme Value Theorem cannot be applied to this interval. The fundamental requirement of continuity over the closed interval is not met, making the theorem inapplicable.

Furthermore, the left endpoint, -3, lies outside the domain of the function, rendering the function undefined at that point. This alone disqualifies the interval from consideration under the Extreme Value Theorem. The theorem explicitly requires the function to be continuous on the closed interval, which implies that the function must be defined at every point within the interval, including the endpoints. The presence of undefined points due to the logarithmic term is a clear violation of this requirement. Therefore, a meticulous examination reveals that the interval [-3, -1] fails to meet the necessary conditions for the Extreme Value Theorem, making its application invalid.

Interval II: [-1, 1]

Considering interval II, [-1, 1], we need to verify the continuity of $y = \frac{\ln(x+2)}{x-2}$. The natural logarithm $\ln(x+2)$ is defined for $x > -2$, and the interval [-1, 1] lies entirely within this domain. The denominator $x-2$ introduces a vertical asymptote at $x = 2$, which is outside the interval [-1, 1]. Thus, the function is continuous on the closed interval [-1, 1]. Since the function is continuous and the interval is closed, the Extreme Value Theorem can be applied to this interval. This satisfies both critical conditions, ensuring the existence of both a maximum and a minimum value for the function within this interval.

The absence of any discontinuities or domain restrictions within the interval [-1, 1] is crucial. The logarithmic component is well-defined, and the potential vertical asymptote at $x=2$ is safely outside the interval's boundaries. This allows the function to maintain a smooth, unbroken graph across the entire interval, fulfilling the continuity requirement of the Extreme Value Theorem. Moreover, the closed nature of the interval, including both endpoints, further solidifies the applicability of the theorem. By confirming both the continuity and the closed interval criteria, we can confidently assert that the function will indeed attain both a maximum and a minimum value within the specified bounds.

Interval III: [1, 3]

Lastly, for interval III, [1, 3], we again assess the function $y = \frac{\ln(x+2)}{x-2}$. As before, $\ln(x+2)$ is defined for $x > -2$, and the interval [1, 3] is within this domain. However, the denominator $x-2$ equals zero when $x = 2$, which lies within the interval [1, 3]. This creates a vertical asymptote at $x = 2$, rendering the function discontinuous at this point. Consequently, the Extreme Value Theorem cannot be applied to this interval because the function is not continuous throughout the closed interval. The presence of a discontinuity within the interval invalidates the theorem's conditions.

The vertical asymptote at $x=2$ is a critical impediment to applying the Extreme Value Theorem on the interval [1, 3]. The function approaches infinity as $x$ approaches 2, creating a break in the graph and violating the continuity requirement. Although the function is defined and continuous on the subintervals [1, 2) and (2, 3], it is not continuous on the closed interval [1, 3] as a whole. This discontinuity prevents the theorem from guaranteeing the existence of maximum and minimum values within the interval. Therefore, the presence of the vertical asymptote within the interval makes the Extreme Value Theorem inapplicable.

Conclusion

In summary, the Extreme Value Theorem can only be applied to interval II, [-1, 1], because it is the only interval where the function $y = \frac{\ln(x+2)}{x-2}$ is continuous on the closed interval. Intervals I and III fail this condition due to domain restrictions and vertical asymptotes, respectively. Therefore, the correct answer is D. II only. Understanding the conditions of the Extreme Value Theorem and the properties of the function is crucial for determining its applicability.

This comprehensive analysis underscores the importance of verifying both the continuity of the function and the closed nature of the interval before attempting to apply the Extreme Value Theorem. A careful examination of the function's domain and potential discontinuities is essential for accurate application of this fundamental theorem in calculus. The ability to correctly identify intervals where the Extreme Value Theorem holds is vital for solving optimization problems and understanding the behavior of functions in various mathematical contexts.