Problem 40 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

Local minimum at \( x = 1, y = 3 \); inflection at \( x = -\sqrt[3]{2} \).

Step 1: Finding the First Derivative

To find the extrema, we need to find the first derivative of the function. The given function is \( y = x^2 + \frac{2}{x} \). We apply the power rule and the derivative of \( \frac{2}{x} \) which is \( -\frac{2}{x^2} \). Thus, the derivative \( y' \) is \( y' = 2x - \frac{2}{x^2} \).

Step 2: Setting the First Derivative to Zero

To find critical points, we set the first derivative equal to zero: \( 2x - \frac{2}{x^2} = 0 \). Simplifying, this gives \( 2x = \frac{2}{x^2} \). Cross-multiplying yields \( 2x^3 = 2 \); thus, \( x^3 = 1 \). Solving this equation gives \( x = 1 \).

Step 3: Finding the Second Derivative

To determine the nature of the critical point and to find inflection points, we calculate the second derivative of the function. The first derivative is \( y' = 2x - \frac{2}{x^2} \). Differentiating again, we obtain \( y'' = 2 + \frac{4}{x^3} \).

Step 4: Evaluating the Second Derivative at the Critical Point

Substitute \( x = 1 \) into the second derivative to check the concavity at this point: \( y''(1) = 2 + \frac{4}{1^3} = 6 \). Since \( 6 > 0 \), the function is concave up at \( x = 1 \), indicating a local minimum.

Step 5: Finding Inflection Points

To find inflection points, we set the second derivative equal to zero: \( 2 + \frac{4}{x^3} = 0 \). Solving gives \( \frac{4}{x^3} = -2 \), leading to \( x^3 = -2 \). Solving for \( x \), we get \( x = -\sqrt[3]{2} \).

Step 6: Determining Function Values at Critical Points

To find the function value at the local minimum \( x = 1 \), substitute into the original function: \( y = 1^2 + \frac{2}{1} = 3 \). To examine the nature of the inflection point \( x = -\sqrt[3]{2} \), compute the function value there to fully understand the curve.

Step 7: Graphing the Function

To showcase the findings, graph the function showing the local minimum at \( x = 1, y = 3 \) and the inflection point around \( x = -\sqrt[3]{2} \). The graph will confirm the concave up region at \( x = 1 \) and capture the curvature change at \( x = -\sqrt[3]{2} \).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that describes how a function changes as its input changes. Essentially, it gives us the rate at which something is increasing or decreasing. For the function given by \( y = x^2 + \frac{2}{x} \), we find its first derivative using known rules.

• Apply the power rule to \( x^2 \), which gives \( 2x \).
• Use the rule for derivatives of fractions on \( \frac{2}{x} \), resulting in \(-\frac{2}{x^2} \).

Combining these, the derivative \( y' \) becomes \( 2x - \frac{2}{x^2} \). This derivative helps determine how the function behaves at various points along the graph, pointing out where the function might reach a peak or valley.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are key because they can indicate where a function has a local maximum or minimum. For the function \( y = x^2 + \frac{2}{x} \), we find the critical points by setting its first derivative to zero:

• \( 2x - \frac{2}{x^2} = 0 \)
• Simplifying yields \( 2x = \frac{2}{x^2} \)
• Cross-multiplying gives \( 2x^3 = 2 \)
• Thus, \( x^3 = 1 \)
• Solving the equation provides \( x = 1 \)

The point \( x = 1 \) is a critical point. It's where the slope of the tangent to the curve is zero, suggesting a potential local extremum.

Inflection Point

An inflection point is where the curve changes its concavity, meaning it shifts from being concave up (shaped like a cup) to concave down (shaped like a cap), or vice versa. To find inflection points, we set the second derivative to zero. The second derivative for our function \( y = x^2 + \frac{2}{x} \) is \( y'' = 2 + \frac{4}{x^3} \). Solving

• \( 2 + \frac{4}{x^3} = 0 \)
• Leads to \( \frac{4}{x^3} = -2 \)
• This gives \( x^3 = -2 \)

Therefore, \( x = -\sqrt[3]{2} \) is an inflection point. At this x-value, the curve changes its shape, altering from concave up to concave down, or vice versa.

Concavity

Concavity tells us about the direction in which a curve opens. If a curve is concave up, it opens upwards like a bowl, indicating the presence of a local minimum. Conversely, if it's concave down, it opens downwards, hinting at a local maximum. By evaluating the second derivative \( y'' = 2 + \frac{4}{x^3} \), we can determine concavity:

• Substituting \( x = 1 \) (a critical point) into \( y'' \) gives us \( y'' = 6 \)
• Since \( 6 > 0 \), the function is concave up at \( x = 1 \)

This confirms a local minimum at the critical point \( x = 1 \). The concavity explains the shape and behavior of a function at and near the critical and inflection points, completing the picture of what the function looks like.