Problem 1 - Calculus Early Transcendentals

Question

Find a number in the closed interval \(\left[\frac{1}{2}, \frac{3}{2}\right]\) such that the sum of the number and its reciprocal is (a) as small as possible (b) as large as possible.

Step-by-Step Solution

Answer

Minimum at 1; maximum at 1/2.

Step 1: Understand the Problem

We are asked to find a number \( x \) in the closed interval \( \left[ \frac{1}{2}, \frac{3}{2} \right] \) such that the sum of the number and its reciprocal, \( x + \frac{1}{x} \), is either as small or as large as possible.

Step 2: Define the Function

Define the function \( f(x) = x + \frac{1}{x} \) which we need to minimize or maximize.

Step 3: Find the Critical Points

To find the critical points, first take the derivative of \( f(x) \):\[f'(x) = 1 - \frac{1}{x^2}\]Set the derivative equal to zero to find critical points:\[1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\]Since \( x \) must be within the interval \( \left[ \frac{1}{2}, \frac{3}{2} \right] \), we only consider \( x = 1 \).

Step 4: Evaluate the Function at Critical and Bound Points

Evaluate \( f(x) \) at key points in the interval. These include the boundaries and the critical point:\[f\left( \frac{1}{2} \right) = \frac{1}{2} + 2 = \frac{5}{2}\]\[f(1) = 1 + 1 = 2\]\[f\left( \frac{3}{2} \right) = \frac{3}{2} + \frac{2}{3} = \frac{13}{6}\]

Step 5: Determine the Minimum and Maximum Values

Compare the values obtained:- \( f(1) = 2 \) is the smallest value.- \( f\left( \frac{5}{2} \right) = \frac{5}{2} \approx 2.5 \) is the largest value.Thus, the minimum occurs at \( x = 1 \) and the maximum at \( x = \frac{1}{2} \).

Key Concepts

Optimization

Optimization in calculus often involves finding the best possible outcome under a certain set of conditions. In this exercise, we are trying to optimize the sum of a number from a specific interval and its reciprocal. Optimization problems usually deal with either minimizing or maximizing a given function. Here, the goal is:

• To find the smallest or largest values of the function within the specified bounds
• To ensure the function is applicable in the given closed interval

By determining the smallest and largest possible values of the sum, we maximize efficiency in decision-making or analysis. Studying optimization techniques sharpens your problem-solving skills and understanding, which can be applied in various fields, from economics to engineering.

Critical Points

Critical points are essential in determining where a function reaches its minimum or maximum value. They occur where the derivative of the function is zero or undefined. In this exercise, the critical points give insight into:

• The potential maximum or minimum points inside the interval
• Helping us to decide if the function increases or decreases around those points

To find critical points for the function in question, we set its derivative equal to zero. Solving this, we identify relevant points within our interval. It is crucial to note that not all critical points in a function's domain are within the provided interval, which means we must verify if our critical points do indeed lie within the interval.

Intervals

When solving for optimization, especially within specific bounds, it is essential to understand the role of intervals. Here, the closed interval \([\frac{1}{2}, \frac{3}{2}]\) restricts us to only consider numbers within this range. Intervals serve several purposes:

• They provide bounds within which the function is analyzed
• They determine which endpoints or critical points need to be evaluated

Since the interval is 'closed', endpoints are considered valid solutions. Therefore, whenever you encounter an interval in similar problems, it's crucial to remember to examine both the boundary points and any critical points that might fall within it.

Derivative

Derivatives are fundamental in calculus for analyzing how a function changes. An understanding of derivatives allows us to understand the function's behavior concerning increasing or decreasing tendencies. In this exercise, calculating the derivative helps:

• Identify where changes occur in the function through critical points
• Determine the direction in which the function is moving (increasing or decreasing)

The derivative of our function, \( f(x) = x + \frac{1}{x} \), is \( f'(x) = 1 - \frac{1}{x^2} \). Solving \( f'(x) = 0 \) allows us to find points where the function stops changing, key to finding optimization results. Solving derivatives accurately ensures effective critical point evaluation, a crucial skill for mastering calculus concepts.