Problem 37 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x \sqrt{8-x^{2}} \end{equation}

Step-by-Step Solution

Answer

Local max at \(x=2\), local min at \(x=-2\), inflection at \(x=0\).

Step 1: Understand the function

The given function is \( y = x \sqrt{8 - x^2} \). This is a product of \( x \) and the square root of another function \( 8 - x^2 \). The domain of the function is determined by the square root component \( \sqrt{8-x^2} \) being non-negative, which implies \( 8 - x^2 \geq 0 \), leading to \( -\sqrt{8} \leq x \leq \sqrt{8} \).

Step 2: Find the derivative

To find critical points for local extrema, we first need to find the derivative of the function. Use the product rule for differentiation. The derivative of \( y = x \sqrt{8 - x^2} \) is \[y' = \sqrt{8 - x^2} + x \left(\frac{-x}{\sqrt{8 - x^2}}\right) = \frac{8 - 2x^2}{\sqrt{8 - x^2}}.\]

Step 3: Solve for critical points

Set the derivative equal to zero to find critical points: \[\frac{8 - 2x^2}{\sqrt{8 - x^2}} = 0.\]This simplifies to \(8 - 2x^2 = 0\), so \(2x^2 = 8\) leading to \(x^2 = 4\). Thus, the critical points are \(x = 2\) and \(x = -2\).

Step 4: Determine the second derivative

Next, find the second derivative to determine the nature of these critical points and find inflection points. The first derivative is \[ y' = \frac{8 - 2x^2}{\sqrt{8 - x^2}}.\]Use the quotient rule to find the second derivative (denoted as \(y''\)): \[ y'' = \frac{d}{dx}\left( \, \frac{8 - 2x^2}{\sqrt{8 - x^2}} \, \right).\]Through calculation, find that \[y'' = \frac{-16x}{(8-x^2)^{3/2}}.\]

Step 5: Evaluate the second derivative at critical points

Evaluate \(y''\) at \(x = 2\) and \(x = -2\).- For \(x = 2\), \(y'' = \frac{-16(2)}{(8-4)^{3/2}} = -8\), which is negative, indicating a local maximum.- For \(x = -2\), \(y'' = \frac{-16(-2)}{(8-4)^{3/2}} = 8\), which is positive, indicating a local minimum.

Step 6: Find inflection points

Inflection points occur where the second derivative changes sign. Set the second derivative equal to zero:\[ \frac{-16x}{(8-x^2)^{3/2}} = 0. \]This occurs only when \(x = 0\). At \(x = 0\), the function changes concavity, verifying an inflection point.

Step 7: Graph the function

The function has the following properties: - Local maximum at \(x = 2\) with \(y = 2\sqrt{4} = 4\).- Local minimum at \(x = -2\) with \(y = -2\sqrt{4} = -4\).- Inflection point at \(x = 0\).Sketch the curve of the function over the interval \([-\sqrt{8}, \sqrt{8}]\) with these points marked, showing the smooth transition and changes in concavity.

Key Concepts

Derivative

Understanding derivatives is essential in calculus as they help to determine the rate of change of a function. For the given function \( y = x \sqrt{8 - x^2} \), finding the derivative involves applying the product rule. This rule is used because the function is the product of \( x \) and a square root component \( \sqrt{8 - x^2} \). The derivative, noted as \( y' \), gives the slope of the tangent to the curve at any point \( x \). In our case, the derivative is calculated as:

• Core derivative: \( y' = \sqrt{8 - x^2} + x \left(\frac{-x}{\sqrt{8 - x^2}}\right) = \frac{8 - 2x^2}{\sqrt{8 - x^2}} \)

This expression helps determine where the slope of the function is zero or undefined, which is crucial for finding critical points and understanding the behavior of the function.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These are potential candidates for local maxima and minima. For our function, to find critical points, we set the derivative \( y' = \frac{8 - 2x^2}{\sqrt{8 - x^2}} \) equal to zero. This simplifies to \( 8 - 2x^2 = 0 \), which gives us the equation \( x^2 = 4 \). Therefore, the critical points are:

• \( x = 2 \)
• \( x = -2 \)

These values are where the function could potentially have local maxima or minima. By further examining these critical points, we can identify their nature through additional tests such as the second derivative test.

Inflection Points

Inflection points are where a function changes concavity. This means the function's curve moves from being concave upwards to concave downwards or vice versa. To identify these points, we look at the second derivative of the function \( y \). For our specific case:

• Second derivative: \( y'' = \frac{-16x}{(8-x^2)^{3/2}} \)

Inflection points occur where \( y'' \) equals zero or changes sign. Solving \( \frac{-16x}{(8-x^2)^{3/2}} = 0 \), we find that the point \( x = 0 \) is an inflection point. Here, the concavity of the function shifts, marking a key change in the behavior of the graph.

Local Extrema

Local extrema are the highest or lowest points within a certain region of a function. These include local maxima and minima which occur at critical points. For the function \( y = x \sqrt{8 - x^2} \), after finding the critical points \( x = 2 \) and \( x = -2 \), we apply the second derivative test to determine the nature of each point:

• At \( x = 2 \), the second derivative \( y'' = -8 \), indicating a local maximum as \( y'' < 0 \).
• At \( x = -2 \), the second derivative \( y'' = 8 \), suggesting a local minimum as \( y'' > 0 \).

Thus, these points symbolize the top and bottom of curvatures, essential in graphing and understanding the overall shape of the function.

Second Derivative Test

The second derivative test is a tool used to determine the concavity of a function at a critical point, which helps in ascertaining whether the point is a local minimum, maximum, or an inflection point. After finding the first derivative and locating critical points, we analyze the second derivative \( y'' \):

• If \( y'' > 0 \) at a critical point, the function is concave up, indicating a local minimum.
• If \( y'' < 0 \) at a critical point, the function is concave down, indicating a local maximum.

For our function, at \( x = 2 \), we found \( y'' = -8 \), showing a local maximum. At \( x = -2 \), with \( y'' = 8 \), there's a local minimum. This test is crucial for graphing functions and comprehending their behavior in the vicinity of critical points.