Problem 41
Question
Finding Absolute Extrema In Exercises \(41-44,\) use a graphing utility to graph the function and find the absolute extrema of the function on the given interval. $$ f(x)=\frac{3}{x-1}, \quad(1,4] $$
Step-by-Step Solution
Verified Answer
The absolute maximum of \(f(x)=\frac{3}{x-1}\) over the interval (1,4] is at \(x = 1\), and the absolute minimum is at \(x = 4\). Note: Although \(x=1\) is not in the interval, we discuss the behavior of the function near this value.
1Step 1: Plot the Function
Firstly, input the function \(f(x)=\frac{3}{x-1}\) into the graphing utility in order to visualize it over the interval (1,4]. It's crucial to focus on the behaviour of the function within the given interval.
2Step 2: Identify Potential Extrema
Next, within the interval (1,4], identify the highest and lowest points on your graph to find the potential absolute extrema. In fact, one must look for local maxima and minima within the interval and the function's value at the endpoint, which is 4 in our case (since including the lower limit will result in an undefined value).
3Step 3: Compare the Values
Finally, compare the value of the function at those points identified in the past step. The absolute maximum is the highest of these values, and the absolute minimum is the lowest. Since the function is continuous on (1,4] and discontinous at \(x = 1\), and decreasing on (1,4], the maximum value should be at \(x = 1\) and the minimum value is at \(x=4\).
Key Concepts
Graphing UtilityContinuity and DiscontinuityLocal Maxima and Minima
Graphing Utility
A graphing utility is an invaluable tool for visualizing mathematical functions and comprehending their behavior over specific intervals. When it comes to finding absolute extrema, which are the highest or lowest values a function takes on a closed interval, graphing utilities simplify the process immensely.
To effectively use a graphing utility, one must input the given function, in our exercise, that's \(f(x)=\frac{3}{x-1}\), and set the appropriate viewing window. This usually involves defining the x-axis and y-axis ranges to capture the interval of interest — in this case, (1,4]. After plotting, the graph reveals not only the general shape and direction of the function but also specific points where the function peaks or dips within the interval.
By zooming in on these areas, students can better identify potential local extremas. It's important, however, to understand that graphing utilities are just one piece of the puzzle, and a thorough analysis of the function's behavior is imperative for accurate conclusions.
To effectively use a graphing utility, one must input the given function, in our exercise, that's \(f(x)=\frac{3}{x-1}\), and set the appropriate viewing window. This usually involves defining the x-axis and y-axis ranges to capture the interval of interest — in this case, (1,4]. After plotting, the graph reveals not only the general shape and direction of the function but also specific points where the function peaks or dips within the interval.
By zooming in on these areas, students can better identify potential local extremas. It's important, however, to understand that graphing utilities are just one piece of the puzzle, and a thorough analysis of the function's behavior is imperative for accurate conclusions.
Continuity and Discontinuity
Understanding the concepts of continuity and discontinuity is essential when examining functions and determining their extremas. A function is continuous at a point if there is no abrupt change in its value, meaning the function is well-defined and follows a smooth path without any jumps, breaks, or holes at that point. Conversely, discontinuity occurs when there is such an interruption in the function's path.
In our textbook exercise, the function \(f(x)=\frac{3}{x-1}\) exhibits discontinuity at \(x = 1\) because the function becomes undefined as \(x\) approaches 1; as a result, there is an asymptotic behavior near this point. The function is, however, continuous on the interval (1,4], which is crucial for determining the absolute extrema. While the absolute maximum cannot occur at a point of discontinuity, the behavior of the function near such points can influence the values of extrema within their vicinity.
In our textbook exercise, the function \(f(x)=\frac{3}{x-1}\) exhibits discontinuity at \(x = 1\) because the function becomes undefined as \(x\) approaches 1; as a result, there is an asymptotic behavior near this point. The function is, however, continuous on the interval (1,4], which is crucial for determining the absolute extrema. While the absolute maximum cannot occur at a point of discontinuity, the behavior of the function near such points can influence the values of extrema within their vicinity.
Local Maxima and Minima
Local maxima and minima represent the peaks and valleys within a specific range of a function. A local maximum is a point where the function takes on a greater value than at any nearby points, while a local minimum is where it takes on a lesser value. These points are critical for finding the absolute extrema on a given interval.
Our function, \(f(x)=\frac{3}{x-1}\), may have local minima or maxima within the interval of (1,4], but we must compare these to the function's value at the interval's endpoints to determine the absolute extrema. It is noteworthy that while local extrema can exist within the open interval, the absolute extrema could also occur at the endpoints if the function's value there is higher or lower than at any other point in the interval.
For the given function, because it is decreasing on (1,4], there are no local maxima within the interval, only a local minimum at \(x=4\). As the function does not exist at \(x=1\) — showing a discontinuity there — we infer that the absolute maximum occurs near this point, but technically not at it due to the open interval. The absolute minimum, in contrast, is located precisely at the closed endpoint, \(x=4\).
Our function, \(f(x)=\frac{3}{x-1}\), may have local minima or maxima within the interval of (1,4], but we must compare these to the function's value at the interval's endpoints to determine the absolute extrema. It is noteworthy that while local extrema can exist within the open interval, the absolute extrema could also occur at the endpoints if the function's value there is higher or lower than at any other point in the interval.
For the given function, because it is decreasing on (1,4], there are no local maxima within the interval, only a local minimum at \(x=4\). As the function does not exist at \(x=1\) — showing a discontinuity there — we infer that the absolute maximum occurs near this point, but technically not at it due to the open interval. The absolute minimum, in contrast, is located precisely at the closed endpoint, \(x=4\).
Other exercises in this chapter
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