Problem 33 - Calculus Early Transcendentals

Question

Use a graphing utility to estimate the absolute maximum and minimum values of \(f \), if any, on the stated interval, and then use calculus methods to find the exact values. \(f(x)=1+\frac{1}{x} ;(0,+\infty)\)

Step-by-Step Solution

Answer

The absolute minimum is 1; there is no finite maximum.

Step 1: Understand the function and interval

We are given the function \( f(x) = 1 + \frac{1}{x} \) and the interval \((0, +\infty)\). This means we need to find the absolute maximum and minimum values of \(f\) on this interval. The function is defined for all \(x > 0\).

Step 2: Use a graphing utility

By graphing the function \(f(x) = 1 + \frac{1}{x}\), we observe that as \(x\) approaches 0 from the right, \(f(x)\) rapidly increases towards infinity. Similarly, as \(x\) approaches infinity, \(f(x)\) approaches 1. From the graph, it appears that the function has an absolute minimum at some finite \(x > 0\) and no finite maximum in \((0, +\infty)\).

Step 3: Calculus method - Find the derivative

To find the critical points of \(f(x)\), we first find the derivative: \(f'(x) = -\frac{1}{x^2}\). Critical points occur where \(f'(x) = 0\) or where the derivative is undefined, but here the derivative can only be undefined at \(x = 0\), which is not in our interval.

Step 4: Analyze the derivative

Since \(f'(x) = -\frac{1}{x^2} < 0\) for all \(x > 0\), the function \(f(x) = 1 + \frac{1}{x}\) is strictly decreasing on \((0, +\infty)\). Therefore, as \(x\) increases, \(f(x)\) approaches 1, the least value it approaches, indicating an absolute minimum at infinity.

Step 5: Conclusion from calculus

The absolute minimum value of \(f(x)\) in the interval \((0, +\infty)\) is 1 as \(x\) approaches infinity. There is no finite absolute maximum on this interval since \(f(x)\) increases indefinitely as \(x\) approaches 0.

Key Concepts

Graphing Utility

A graphing utility is a helpful tool that uses computer software, graphing calculators, or online graphing tools to visualize mathematical functions. In this exercise, we used a graphing utility to assess the behavior of the function \( f(x) = 1 + \frac{1}{x}\) over the interval \((0, +\infty)\). By inputting the function into the graphing utility, we can generate a visual representation of \(f(x)\).

This visualization helps in estimating the absolute maximum and minimum values over the given interval. When we graph \( f(x)\), we notice how as \( x \) approaches 0 from the positive side, the function value spikes upwards towards infinity. Conversely, as \( x \) becomes very large, or approaches infinity, \( f(x)\) stabilizes towards the value 1. This graphical approach provides a preliminary understanding of the function's behavior, guiding us in analyzing the function further using calculus methods.

Derivative

The derivative of a function is a fundamental tool in calculus that measures the rate at which a function changes. It is essentially the slope of the function at any given point. For the function \( f(x) = 1 + \frac{1}{x} \), we calculate the derivative to find critical points, where the function might have local maxima or minima.

To find the derivative, we apply the basic rules of differentiation. The derivative of \( f(x) \) is \( f'(x) = -\frac{1}{x^2} \). This derivative tells us how the function \( f(x) \) is changing at each point for \( x > 0 \). Since \( f'(x) = -\frac{1}{x^2} \) is always less than zero in our interval (because \( x^2 \) is always positive for \( x > 0 \)), it indicates that the function is decreasing throughout this interval.

Critical Points

Critical points of a function exist where its derivative is zero or undefined. These points help identify where the function reaches local minimas or maximas. In the context of the function \( f(x) = 1 + \frac{1}{x} \), the derivative \( f'(x) = -\frac{1}{x^2} \) provides no critical points within the interval \((0, +\infty)\) since \( f'(x) \) is never zero.

At the boundary of our interval, \( x = 0 \), the function is undefined, hence not considered a critical point within our specified range. Importantly, this lack of critical points within the interval results from the fact that the entire interval shows a monotonic behavior, that is, the function is strictly decreasing without turning points that create peaks or valleys.

Interval Analysis

Interval analysis involves evaluating the behavior of a function over a given interval to determine various attributes like continuity and extrema. For the function \( f(x) = 1 + \frac{1}{x} \) over the interval \((0, +\infty)\), we use this analysis to investigate and confirm our findings.

Since \( f'(x) = -\frac{1}{x^2} < 0 \) for all \( x > 0 \), the function is strictly decreasing. As \( x \) increases toward infinity, \( f(x) \) approaches 1, signifying that 1 is the lowest value or absolute minimum over this interval. However, due to the function's increase towards infinity as \( x \) approaches 0, there is no absolute maximum within the interval.

• This confirms our earlier graphical insight and calculus findings.
• Interval analysis gives a concrete framework for understanding how this behavior unfolds within the specified domain, leading to accurate interpretations of maxima and minima.