Problem 35 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x^{2 / 3}\left(\frac{5}{2}-x\right) \end{equation}

Step-by-Step Solution

Answer

The absolute maximum is \( (1, \frac{3}{2}) \) with no inflection points.

Step 1: Find the derivative

To identify critical points, which may indicate local extreme points, we first find the derivative of the function. Let's compute the derivative of the function \( y = x^{2/3} \left( \frac{5}{2} - x \right) \) using the product rule: \( u = x^{2/3} \), \( v = \frac{5}{2} - x \). Derivatives: \( u' = \frac{2}{3} x^{-1/3} \), \( v' = -1 \). Applying product rule: \( y' = uv' + vu' = x^{2/3}(-1) + \left( \frac{5}{2} - x \right) \left( \frac{2}{3} x^{-1/3} \right) \). Simplify: \( y' = -x^{2/3} + \frac{5}{3} x^{-1/3} - \frac{2}{3} x^{2/3} \). Final form: \( y' = -\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3} \).

Step 2: Set the derivative equal to zero

Next, set the derivative \( y' = -\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3} \) to zero to find critical points: \( 0 = -\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3} \). This simplifies to \( 0 = x^{-1/3} (\frac{5}{3} - \frac{5}{3} x) \), giving \( x - 1 = 0 \). Thus \( x = 1 \). We also consider where the derivative may be undefined; here, it's undefined at \( x = 0 \). Therefore, the critical points are \( x = 0 \) and \( x = 1 \).

Step 3: Test for local extrema

To determine the nature of these critical points, we can use the Second Derivative Test. Compute the second derivative: starting from \( y' = -\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3} \), let's differentiate: \( y'' = -\frac{10}{9} x^{-1/3} - \frac{5}{9} x^{-4/3} \). Evaluating at \( x = 1 \): \( y''(1) = -\frac{10}{9} - \frac{5}{9} = -\frac{15}{9} \), which is negative, indicating a local maximum at \( x = 1 \).

Step 4: Identify absolute extrema

Since the function is defined for \( x \geq 0 \), and only a local maximum is found, check the endpoints or limits. As \( x \rightarrow 0^+ \), \( y \approx \frac{5}{2} x^{2/3} \), decreasing from a small positive value, and as \( x \rightarrow \infty \), \( y \rightarrow -\infty \) quickly. The absolute maximum occurs at \( x = 1 \) with \( y(1) = x^{2/3}(\frac{5}{2}-x) = \frac{3}{2} \).

Step 5: Determine inflection points

An inflection point occurs where the second derivative changes sign. For \( y'' = -\frac{10}{9} x^{-1/3} - \frac{5}{9} x^{-4/3} \), consider sign changes: \( y'' \) is negative for \( x > 0 \), meaning it doesn't change sign, indicating no inflection points.

Step 6: Graph the function

Plot the function with identified key features. At \( x = 0 \), the derivative is undefined, while \( x = 1 \) is a local and absolute maximum with \( y(1) = \frac{3}{2} \). For large \( x \), \( y \rightarrow -\infty \). Begin graph just above the origin, peaking at 1, and declining into negative values.

Key Concepts

Derivative Calculations

To find critical points and understand the behavior of a function, the first step is to find the derivative. The derivative gives us the rate at which the function changes, shedding light on where the function increases or decreases. In our function, \(y = x^{2/3} \left( \frac{5}{2} - x \right)\), we apply the product rule to find the derivative. The function is composed of two parts:

• \(u = x^{2/3}\)
• \(v = \frac{5}{2} - x\)

Calculating each part's derivative:

• \(u' = \frac{2}{3} x^{-1/3}\)
• \(v' = -1\)

The product rule is applied as follows: \(y' = uv' + vu' = x^{2/3}(-1) + \left( \frac{5}{2} - x \right) \left( \frac{2}{3} x^{-1/3} \right)\). This simplifies to: \(y' = -x^{2/3} + \frac{5}{3} x^{-1/3} - \frac{2}{3} x^{2/3}\),leading to the final form:\(y' = -\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3}\). Here, the derivative is a crucial tool to identify critical points, where changes in direction, such as maxima or minima, might occur.

Second Derivative Test

Once you have the first derivative and the critical points, the second derivative test helps determine the nature of these critical points. The second derivative, denoted as \(y''\), informs us about the concavity of the function, whether it curves upwards or downwards. For our function, the second derivative is: \(y'' = -\frac{10}{9} x^{-1/3} - \frac{5}{9} x^{-4/3}\). Here’s how the test works:

• A positive second derivative (\(y'' > 0\)) indicates the function is concave up, suggesting a local minimum.
• A negative second derivative (\(y'' < 0\)) indicates the function is concave down, suggesting a local maximum.

To apply this to our critical point at \(x = 1\): \(y''(1) = -\frac{10}{9} - \frac{5}{9} = -\frac{15}{9}\). Since this value is negative, it shows a local maximum at \(x = 1\). The second derivative test simplifies understanding complex behavior in graphs by categorizing the nature of points.

Critical Points

Critical points occur where the derivative equals zero or is undefined. These points are crucial because they often correspond to local maxima, minima, or inflection points. For the function \(y = x^{2/3} \left( \frac{5}{2} - x \right)\), the derivative:\(-\frac{5}{3} x^{2/3} + \frac{5}{3} x^{-1/3}\) simplifies to zero at critical points.By setting the derivative to zero,\(0 = x^{-1/3} \left( \frac{5}{3} - \frac{5}{3} x \right)\),we solve for:

• \(x = 1\)

We must also consider where the derivative is undefined, which in this case, occurs at

• \(x = 0\).

This gives us our critical points: \(x = 0\) and \(x = 1\). Each of these critical points provides valuable insights into the shape and direction of the function graph. However, further testing, such as a second derivative check, is necessary to understand if they are maxima, minima, or neither.

Local and Absolute Extrema

Analyzing critical points allows us to discern local and absolute extrema. Local extrema can be seen as 'peaks' and 'valleys' in a graph over a given interval, while absolute extrema are the highest or lowest points over the entire domain.In our example, the focus is primarily on non-negative \(x\) values. With the second derivative test, we found that \(x = 1\) is a local maximum since the second derivative is negative there:

• \(y(1) = 1^{2/3} \left( \frac{5}{2} - 1 \right) = \frac{3}{2}\).

This point is also an absolute maximum given that as \(x\) grows, the function value decreases towards negative infinity.At extreme values like \(x = 0\):

• The function approaches a small positive value due to the \(x^{2/3}\) term.

Understanding these concepts helps predict and interpret function behavior in various scenarios, facilitating deeper comprehension of calculus.

Graphing Functions

Graphing a function visually represents the mathematical relationship and highlights key characteristics like extrema and concavity. The function \(y = x^{2/3} \left( \frac{5}{2} - x \right)\) lives within domain constraints- primarily \(x \geq 0\) in this case. Plotting begins just above the origin and quickly ascends towards \(x = 1\), where it achieves a peak described as both a local and absolute maximum. Key stages in the graph include:

• Beginning near the origin due to the \(x^{2/3}\) effect, creating a shallow initial slope.
• Reaching the maximum value \(y(1) = \frac{3}{2}\) at \(x = 1\).
• Declining sharply beyond the maximum leading towards negative infinity as \(x\) grows.

Visualizing the function offers a comprehensive understanding of the mathematical concepts and enhances our ability to connect calculus principles with concrete examples.