Problem 29 - Thomas Calculus

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 positive number is 1.

Step 1: Understanding the Problem

We are asked to find a positive number, say \( x \), such that the sum of the number and its reciprocal, \( x + \frac{1}{x} \), is minimized. Our goal is to find the value of \( x \) that minimizes this expression.

Step 2: Set up the Function

Define the function \( f(x) = x + \frac{1}{x} \). We want to find the minimum value of this function for \( x > 0 \).

Step 3: Differentiate the Function

To find the minimum, differentiate \( f(x) \) with respect to \( x \). The derivative is \( f'(x) = 1 - \frac{1}{x^2} \).

Step 4: Set the Derivative to Zero

To find the critical points, set \( f'(x) = 0 \). That gives us the equation \( 1 - \frac{1}{x^2} = 0 \). Solving this, we get \( x^2 = 1 \), which implies \( x = 1 \) (since \( x \) is positive).

Step 5: Verify the Minimum with the Second Derivative

Find the second derivative: \( f''(x) = \frac{2}{x^3} \). At \( x = 1 \), \( f''(1) = 2 \) which is positive. This confirms that \( x = 1 \) is a point of local minima.

Step 6: Conclusion

Since \( x = 1 \) minimizes \( f(x) \) and the second derivative test confirms it, the positive number that makes the sum of it and its reciprocal smallest is 1.

Key Concepts

Critical Points

To understand optimization problems like this one, finding the critical points is crucial. Critical points are values where the function's rate of change (derivative) is either zero or undefined. These points are potential locations for local maxima, minima, or saddle points.
In this exercise, we identified a critical point by setting the derivative of the function, \( f(x) = x + \frac{1}{x} \), equal to zero. This resulted in the equation \( 1 - \frac{1}{x^2} = 0 \). Solving this gives \( x = 1 \), indicating a critical point at \( x = 1 \).
Finding critical points is the first step in determining where a function might achieve its smallest or largest values within its domain.

Derivative

Derivatives are foundational in calculus for understanding how functions change. When we differentiate a function, we obtain its derivative, which provides the slope of the tangent line at any point of the function. This slope tells us the rate at which the function value is changing.
In this problem, the function \( f(x) = x + \frac{1}{x} \) was differentiated to yield \( f'(x) = 1 - \frac{1}{x^2} \). Each term represents a different part of the function's behavior:

• \( 1 \) is constant, indicating a flat slope component.
• \( -\frac{1}{x^2} \) shows how quickly the reciprocal term decreases.

The derivative is set to zero to find critical points, revealing how changes in \( x \) impact the overall function value.

Minimization

Minimization is the process of finding the smallest possible value of a function over its domain. In the context of the given exercise, the goal was to find the smallest possible sum of a number and its reciprocal.
This is achieved by evaluating the function \( f(x) = x + \frac{1}{x} \) at potential critical points, such as the one found at \( x = 1 \). If this point lowers the function value more than any nearby points, it indicates a local minimum.
Optimization practices like these help solve real-world questions where reducing costs or maximizing efficiency is crucial.

Second Derivative Test

The second derivative test is used to confirm whether a critical point is indeed a minima or maxima. It involves taking the second derivative of the function and checking its sign at the critical point:

• If \( f''(x) > 0 \), the function is concave up, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.

In this problem, the second derivative of \( f(x) = x + \frac{1}{x} \) is \( f''(x) = \frac{2}{x^3} \). At \( x = 1 \), \( f''(1) = 2 \), which is positive, confirming that the function is concave up and \( x = 1 \) is indeed a point of local minima.
This test is a powerful tool for verifying optimization results by analyzing the curvature of the function.