Problem 29 - University Calculus: Early Transcendentals

Question

Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible.

Step-by-Step Solution

Answer

The number is 1.

Step 1: Understand the Problem

We need to find a positive number, say \(x\), such that the sum of the number and its reciprocal, \( x + \frac{1}{x} \), is the smallest possible. This is an optimization problem.

Step 2: Set Up the Function

Define the function \( f(x) = x + \frac{1}{x} \). We need to minimize this function for \(x > 0\).

Step 3: Find the Derivative

Calculate the derivative of \( f(x) \). The derivative is\[ f'(x) = 1 - \frac{1}{x^2}. \]

Step 4: Find Critical Points

Set the derivative equal to zero to find critical points:\[ 1 - \frac{1}{x^2} = 0 \]which leads to \( x^2 = 1 \). Thus, \( x = 1 \) (since \(x\) is positive).

Step 5: Verify Minimum Point

To ensure this is a minimum, check the second derivative:\[ f''(x) = \frac{2}{x^3}. \]Since \(f''(1) = 2 > 0\), the function has a local minimum at \(x = 1\).

Step 6: Conclusion

The function \( f(x) = x + \frac{1}{x} \) is minimized when \(x = 1\). Therefore, the positive number for which the sum of it and its reciprocal is the smallest is 1.

Key Concepts

Critical Points

In calculus, critical points are values in the domain of a function where its derivative is zero or undefined. These points are crucial in identifying where a function reaches its local maximum or minimum. For the function given in the problem, we define it as

1. \( f(x) = x + \frac{1}{x} \)

To find the critical points, we first calculate the derivative \( f'(x) = 1 - \frac{1}{x^2} \). Critical points occur where \( f'(x) = 0 \) or is undefined.

• Setting \( f'(x) = 0 \), we solve \( 1 - \frac{1}{x^2} = 0 \), leading to \( x^2 = 1 \).
• Since we are only considering positive numbers, we find \( x = 1 \).

Critical points help us determine where to check for potential minima or maxima of our function, which are important for optimization problems.

Derivative

The derivative is a fundamental tool in calculus that measures how a function changes as its input changes. It essentially provides the slope of the tangent line to the graph of the function at any point.
In our optimization problem, the function whose behavior we're analyzing is \( f(x) = x + \frac{1}{x} \). To analyze its changes, we take its derivative:

• Calculate \( f'(x) = 1 - \frac{1}{x^2} \).

The derivative \( f'(x) \) shows the rate at which \( f(x) \) changes with \( x \).
Using the derivative helps us find critical points and identify potential local minima or maxima, crucial steps in determining the smallest possible sum in our problem.

Second Derivative Test

The second derivative test is a technique to determine whether a critical point of a function is a local minimum, local maximum, or a point of inflection.
In this specific optimization problem, we've calculated the second derivative to confirm that the critical point is a local minimum.

• The function \( f(x) = x + \frac{1}{x} \) yields a second derivative \( f''(x) = \frac{2}{x^3} \).
• We evaluate this at the critical point \( x = 1 \), resulting in \( f''(1) = 2 \).

Since \( f''(1) = 2 \) is greater than zero, it indicates that the graph of \( f(x) \) is concave up at \( x = 1 \), confirming a local minimum.
This affirmation with the second derivative test ensures that we've located the smallest value for the given problem.