Problem 29 - Thomas Calculus in SI Units

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 that minimizes the sum of it and its reciprocal is 1.

Step 1: Define the problem

Let's denote the positive number we're looking for as \( x \). The problem asks us to find a positive number such that the sum of it and its reciprocal is minimized. This means we want to minimize the function \( f(x) = x + \frac{1}{x} \) for \( x > 0 \).

Step 2: Find the derivative

To find where the function reaches its minimum, we first need the derivative of the function: \( f(x) = x + \frac{1}{x} \). The derivative is \( f'(x) = 1 - \frac{1}{x^2} \).

Step 3: Set the derivative to zero

To find the critical points, set the derivative \( f'(x) = 1 - \frac{1}{x^2} = 0 \). Solving this gives: \[ 1 = \frac{1}{x^2} \] or \[ x^2 = 1 \] which implies \[ x = 1 \] since \( x \) is positive.

Step 4: Verify it's a minimum

Use the second derivative test to confirm that \( x = 1 \) is a minimum. Compute the second derivative:\( f''(x) = \frac{2}{x^3} \). At \( x = 1 \), \( f''(1) = 2 \), which is positive, indicating a local minimum.

Step 5: Conclusion

Therefore, the positive number \( x = 1 \) is where the sum of the number and its reciprocal is minimized. At \( x = 1 \), the sum \( f(1) = 1 + \frac{1}{1} = 2 \).

Key Concepts

Critical Points

In calculus, critical points of a function are where the function's derivative is either zero or undefined. These points are significant because they can indicate where a function achieves its local minimums or maximums.
Understanding critical points is crucial for optimization problems, like finding the minimum value of a function. They help us determine intervals where a function might change its behavior.
To locate critical points, take the derivative of the function and solve for when this derivative is equal to zero. This tells us the points where the function might change direction, either peaking or dipping. In our problem, for example, setting the derivative \( f'(x) = 1 - \frac{1}{x^2} \) to zero helps us find that the critical point is \( x = 1 \). But remember, finding a critical point is only the first step. More tests need to be conducted to confirm whether these points are minimums or maximums.

Derivative Test

The derivative test helps determine whether a critical point is a minimum, maximum, or neither. Once you've calculated critical points, use the derivative test by evaluating the first derivative.
Here's how you do it:

• Find the derivative of your function. For instance, the function \( f(x) = x + \frac{1}{x} \) has a derivative of \( f'(x) = 1 - \frac{1}{x^2} \).
• Find critical points by setting \( f'(x) \) to zero. We found \( x = 1 \) in our example.
• Determine if the function changes its rate of increase or decrease before and after these points. If \( f'(x) \) changes from positive to negative at a point, it's a local maximum. If it changes from negative to positive, you found a local minimum.

In our problem, the derivative does not need to change signs because \( f'(x) = 0 \) was only checked at a specific critical point. Instead, the second derivative test was applied to confirm the kind of critical point.

Reciprocal Function

A reciprocal function is expressed as the inverse of another function. In simplest terms, if \( y = f(x) \), then the reciprocal is \( y = \frac{1}{f(x)} \).
This function is interesting in optimization problems. Like in our example, we had to minimize \( x + \frac{1}{x} \). The term \( \frac{1}{x} \) represents the reciprocal of \( x \). It is significant here since it introduces a component that behaves differently from \( x \).
For values of \( x \) approaching zero, \( \frac{1}{x} \) grows very large. Conversely, as \( x \) increases, \( \frac{1}{x} \) shrinks toward zero, making understanding of reciprocal relationships important for predicting the behavior of such functions.

Second Derivative Test

The second derivative test is a handy tool in verifying whether a critical point is a minimum or maximum. It complements the first derivative test.
Here's how we use the second derivative test:

• Calculate the second derivative of your function.
• Evaluate the second derivative at each critical point.
• If \( f''(x) \) is positive at a critical point, the function has a local minimum there.
• If \( f''(x) \) is negative, there's a local maximum.

In our optimization problem, after finding the critical point \( x = 1 \), we calculated the second derivative: \( f''(x) = \frac{2}{x^3} \). Evaluating at \( x = 1 \), we find \( f''(1) = 2 \), which is positive. Hence, \( x = 1 \) provides a local minimum, proving it to be the point where our original function is minimized.