Problem 40 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x^{2}+\frac{2}{x} \end{equation}

Step-by-Step Solution

Answer

Local minimum at \( (1, 3) \); inflection point at \( (-\sqrt[3]{2}, 0) \).

Step 1: Find the First Derivative

To identify the critical points, start by finding the first derivative of the function. Given \( y = x^2 + \frac{2}{x} \), we use the power rule and the derivative of \( \frac{1}{x} \). The first derivative is: \( y' = 2x - \frac{2}{x^2} \).

Step 2: Set the First Derivative to Zero

To find critical points where local extrema may occur, set \( y' = 0 \). Solving the equation \( 2x - \frac{2}{x^2} = 0 \) gives \( 2x = \frac{2}{x^2} \). Multiply by \( x^2 \) to eliminate the fraction: \( 2x^3 = 2 \). Simplifying, we have \( x^3 = 1 \), leading to \( x = 1 \).

Step 3: Find Second Derivative

Next, find the second derivative to apply the second derivative test and find inflection points. The second derivative of the function \( y = x^2 + \frac{2}{x} \) is \( y'' = 2 + \frac{4}{x^3} \).

Step 4: Apply Second Derivative Test

Use the second derivative test to determine the nature of the critical point at \( x = 1 \). Evaluate \( y''(1) = 2 + \frac{4}{1^3} = 6 \). Since \( y''(1) > 0 \), the function has a local minimum at \( x = 1 \).

Step 5: Check for Inflection Points

Inflection points occur where \( y'' = 0 \) or changes sign. Set the second derivative to zero: \( 2 + \frac{4}{x^3} = 0 \). Solving, \( \frac{4}{x^3} = -2 \) simplifies to \( x^3 = -2 \), giving \( x = -\sqrt[3]{2} \). This is the x-coordinate of the inflection point.

Step 6: Identify Absolute Extrema

Since the function is undefined at \( x = 0 \), consider \( x \to 0^+ \) and \( x \to 0^- \) to assess bounds. As \( x \to \infty \) or \( x \to -\infty \), \( y \to \infty \). The function has no absolute maximum or minimum on its entire domain (except at critical points), because the ranges tend to infinity.

Key Concepts

Critical Points

Critical points are essential in calculus because they indicate where the function's slope is zero or undefined, signaling possible local maxima, minima, or points of inflection. For the function given, the first derivative is used to find these points. It involves setting the derivative to zero, resulting in candidate points like where changes may occur in the function's behavior.

Finding critical points requires:

• Taking the derivative of the function, which represents its rate of change.
• Setting the derivative equal to zero to find potential extreme points.
• Ensuring the derivative is defined at those points, or noting where it is not, as they could also represent critical points.

In the exercise, setting the first derivative to zero and solving the equation provided a critical point at \( x = 1 \). This point provides clues on where the function might change its increasing or decreasing trend.

First Derivative

The first derivative of a function provides us with critical information about the function's slope. It allows us to locate critical points and understand the function's increasing or decreasing nature. For the function \( y = x^2 + \frac{2}{x} \), the first derivative \( y' \) was computed as \( y' = 2x - \frac{2}{x^2} \).

When analyzing the first derivative:

• A positive first derivative indicates that the function is increasing.
• A negative first derivative signifies a decreasing function.
• A zero derivative suggests a potential local maximum or minimum, also pointing us towards critical points.

In this example, by setting the first derivative equal to zero, \( 2x - \frac{2}{x^2} = 0 \), it was discovered that \( x = 1 \) is a critical point, a pivotal part of determining extreme points on the graph.

Second Derivative

The second derivative of a function offers insights on the curvature and concavity of the function. It plays a crucial role in identifying the nature of critical points through the second derivative test. For the provided function, the second derivative is \( y'' = 2 + \frac{4}{x^3} \).

Key aspects of using the second derivative include:

• Determining concavity: If \( y'' > 0 \), the function is concave up; if \( y'' < 0 \), concave down.
• Applying the second derivative test at critical points to confirm whether they are local maxima or minima.
• Finding inflection points where the function changes concavity.

In the case of \( x = 1 \), calculating \( y''(1) = 6 \) affirmed the presence of a local minimum due to its positive value, indicating the curve is concave up at this critical point.

Local Minimum

A local minimum occurs at a point where the function reaches its lowest value in a neighborhood around that point, but not necessarily throughout its entire domain. It is confirmed using both the first and second derivative tests if applicable.

In this exercise, after finding the critical point \( x = 1 \), the second derivative test was applied, yielding \( y''(1) = 6 \), which is greater than zero. This means the graph is concave upward at \( x = 1 \), affirming it as a local minimum. Such confirmations are vital, especially in calculus, as they help us predict behavior and identify valuable or harmful points in real-world applications.

Inflection Point

Inflection points signify where a function's curve changes from concave upward to concave downward or vice versa. They are crucial in comprehending a function's shape and are identified by where the second derivative equals zero or changes sign.

The process involves:

• Setting the second derivative to zero and solving for \( x \).
• Ensuring the derivative changes sign around these values.

For \( y'' = 2 + \frac{4}{x^3} \), setting \( y'' = 0 \) and solving provided \( x = -\sqrt[3]{2} \) as an inflection point. This indicates a change in concavity, illustrating how the function's shape is adjusted at this point, which can be highly informative in graphing and understanding curves.