Problem 39 - Calculus of a Single Variable

Question

Using the Second Derivative Test In Exercises \(31-42\) , find all relative extrema. Use the Second Derivative Test where applicable. $$ f(x)=x+\frac{4}{x} $$

Step-by-Step Solution

Answer

The function \(f(x) = x + \frac{4}{x}\) has a local minimum at \(x = 2\) and a local maximum at \(x = -2\).

Step 1: Differentiation

First we will find the derivative of the function. For the function \(f(x) = x + \frac{4}{x}\), the derivative using power and constant rules is \(f'(x) = 1 - 4x^{-2}\). This simplifies to \(f'(x) = 1 - \frac{4}{x^2}\).

Step 2: Find Critical Points

Critical points occur where the derivative is 0 or undefined. To find them, we set \(f'(x) = 1 - \frac{4}{x^2}\) equal to 0 and solve for x, which gives the critical points \(x = 2, -2\). The derivative is undefined when \(x = 0\), but as the original function is also undefined there, this is not a critical point.

Step 3: Second Derivative

We find the second derivative to apply the Second Derivative Test. Thus, \(f''(x) = 8x^{-3}\), which can be written as \(f''(x) = \frac{8}{x^3}\).

Step 4: Second Derivative Test

To classify the critical points, we plug them into the second derivative. With \(x = 2\), the second derivative becomes positive, indicating a local minimum. With \(x = -2\), the second derivative becomes negative, indicating a local maximum. Therefore, the function has a local minimum at \(x = 2\) and a local maximum at \(x = -2\).

Key Concepts

Critical Points in Calculus

In the realm of calculus, critical points are essential for understanding the behavior of functions. Simply put, critical points are the x-values where the function's derivative is either zero or undefined. These points can potentially be where the function reaches a local maximum or minimum, or where there might be a point of inflection.

When analyzing the function f(x) = x + \(\frac{4}{x}\), we find its derivative and set it equal to zero. The critical points here, x = 2 and x = -2, are where the derivative f'(x) = 1 - \(\frac{4}{x^2}\) is zero. Importantly, critical points are not guaranteed to be points of maxima or minima; they are simply the 'candidates' which need further testing with methods like the Second Derivative Test or the First Derivative Test.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function. It is a fundamental tool in calculus used to analyze the rate of change of quantities. When differentiating functions like f(x) = x + \(\frac{4}{x}\), we apply differentiation techniques such as the power rule, the constant rule, and the quotient rule.

To find the first derivative of this particular function, f'(x) = 1 - \(4x^{-2}\), we use the power rule to handle the term x, which is written as x^1, and the constant multiplication rule for the term \(\frac{4}{x}\), which is seen as 4x^{-1}. The understanding and proper application of these rules allow us to accurately find critical points and predict the behavior of functions at these points.

Local Extrema

Determining where a function takes on its highest or lowest values within a certain interval is crucial for many applications in science, engineering, and mathematics. These points are known as local extrema. A local minimum is where the function has the lowest value in the nearby region, while a local maximum is where the function has the highest value in that vicinity.

To find local extrema, we use the Second Derivative Test as we did for the function f(x) = x + \(\frac{4}{x}\). After finding the second derivative, f''(x) = \(\frac{8}{x^3}\), we evaluate it at the critical points. If the second derivative is positive at a critical point, like it is at x = 2, it indicates a local minimum. Conversely, if it's negative, as it is at x = -2, it indicates a local maximum. This test provides a quick way to classify critical points without examining the original function over an interval.