Problem 28 - Applied Calculus

Question

Find the exact global maximum and minimum values of the function. The domain is all real numbers unless otherwise specified. $$f(x)=x+1 / x \text { for } x>0$$

Step-by-Step Solution

Answer

The global minimum is 2 at \( x=1 \). There is no global maximum.

Step 1: Understand the Function

The function given is \( f(x) = x + \frac{1}{x} \) and it's defined for \( x > 0 \). We need to find its global maximum and minimum values in this domain.

Step 2: Find the Derivative

To find critical points, we first compute the derivative of the function. The derivative \( f'(x) \) is found by differentiating \( f(x) = x + \frac{1}{x} \). Using the power rule and chain rule, we have \( f'(x) = 1 - \frac{1}{x^2} \).

Step 3: Set the Derivative to Zero

To find critical points, set the derivative to zero: \( 1 - \frac{1}{x^2} = 0 \). Solving this equation gives \( x^2 = 1 \), so \( x = 1 \) (since \( x > 0 \)).

Step 4: Determine Critical Point Nature

Evaluate the second derivative to determine if \( x=1 \) is a max or min. Compute \( f''(x) = \frac{2}{x^3} \). Since \( f''(1) = 2 > 0 \), \( x=1 \) is a point of local minimum.

Step 5: Evaluate Function at Boundary and Critical Points

Calculate \( f(x) \) at critical point: \( f(1) = 1 + \frac{1}{1} = 2 \). As \( x \to 0^+ \), \( f(x) \to +\infty \), and as \( x \to \infty, f(x) \to \infty \).

Step 6: Conclude the Results

Since the function approaches infinity as \( x \) approaches 0 from the right or \( x \to \infty \), and the minimum value of 2 is found at \( x=1 \), the global minimum is 2, attained at \( x=1 \). No global maximum exists as \( f(x) \to \infty \) as \( x \to 0^+ \) and \( x \to \infty \).

Key Concepts

Critical Points

In calculus, critical points are a fundamental concept for understanding the behavior of a function. A critical point occurs where the function's derivative equals zero or is undefined. These points are crucial because they can indicate where the function might take a turn, either reaching a peak (maximum) or a trough (minimum). For a function \( f(x) \), the critical points are found by solving \( f'(x) = 0 \) or where \( f'(x) \) does not exist.
In the original exercise, the function is \( f(x) = x + \frac{1}{x} \). To find the critical points, the derivative is set to zero: \( 1 - \frac{1}{x^2} = 0 \). Solving this, we find that \( x = 1 \) is a critical point, which is important for further analysis.

• Critical Points help in locating potential maxima or minima.
• They play a key role in graph analysis and optimization problems.

Derivative

The derivative of a function is a measure of how the function's output changes as its input changes. It is a core concept in calculus, often representing the slope of the tangent line to the curve of a function at a given point. By finding the derivative, we can understand the rate of change and identify critical points.
To compute the derivative of our function \( f(x) = x + \frac{1}{x} \), we use basic differentiation rules. The derivative is \( f'(x) = 1 - \frac{1}{x^2} \). This expression helps us determine where the slope of the tangent to our function is zero or undefined, leading us to critical points.
Derivatives are foundational for:

• Identifying increasing or decreasing intervals of functions.
• Understanding and finding the exact slope at any point on a function.

Second Derivative Test

The second derivative test is a technique used to classify the nature of critical points found in a function. After identifying a critical point, the second derivative of the function is evaluated at this point. The result helps determine whether the critical point is a local maximum, local minimum, or neither.
For the function \( f(x) = x + \frac{1}{x} \), we first found the critical point at \( x = 1 \). The second derivative is \( f''(x) = \frac{2}{x^3} \). By evaluating the second derivative at \( x = 1 \), \( f''(1) = 2 \), we see that it is positive. Thus, according to the second derivative test, \( x = 1 \) is a point of local minimum.
Second derivative test helps:

• Confirm the nature of critical points efficiently.
• Determine concavity and curvature of the graph at a certain point.

Global Maximum and Minimum

In optimization problems, identifying the global maximum and minimum of a function is essential, especially in various applied contexts. A global maximum or minimum is the highest or lowest output a function can produce across its entire domain.
For our function \( f(x) = x + \frac{1}{x} \), we evaluated the behavior at different critical points and the boundaries. With \( x > 0 \), as \( x \to 0^+ \) or \( x \to \infty \), \( f(x) \) tends to infinity. This means there is no global maximum because the function keeps increasing without bound. However, at \( x = 1 \), we found the global minimum value to be 2.
This concept is critical in:

• Solving real-world problems where extremes are needed.
• Ensuring decisions are made based on the best possible outcome.