Problem 38 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Local maximum at \((0, 2^{3/2})\), inflection points at \((1, 1)\) and \((-1, 1)\), with x-intercepts at \((\pm \sqrt{2}, 0)\).

Step 1: Find the First Derivative

To find the local extreme points, we start by finding the first derivative of the function \[ y = (2 - x^2)^{3/2} \] using the chain rule. Let \[ u = 2 - x^2 \] so that \[ y = u^{3/2} \]. The derivative of \(y\) with respect to \(x\) is:\[ \frac{dy}{dx} = \frac{d}{du}(u^{3/2}) \cdot \frac{du}{dx} \]\[ \frac{d}{du}(u^{3/2}) = \frac{3}{2}u^{1/2} \] and \[ \frac{du}{dx} = -2x \]. Thus, \[ \frac{dy}{dx} = \frac{3}{2}(2-x^2)^{1/2}(-2x) = -3x(2-x^2)^{1/2} \].

Step 2: Find Critical Points

Set the first derivative equal to zero to find critical points:\[ -3x(2-x^2)^{1/2} = 0 \]This equation is zero when \(x = 0\) and also when the inside of the square root is zero, i.e., when \(2-x^2 = 0\). Solving \(2-x^2 = 0\) gives \(x = \pm\sqrt{2}\). Therefore, the critical points are at \(x = 0, x = \sqrt{2}, \text{and } x = -\sqrt{2}\).

Step 3: Find the Second Derivative

To determine the concavity and find inflection points, we take the second derivative of \[ y = (2-x^2)^{3/2} \]. Using the first derivative \( \frac{dy}{dx} = -3x(2-x^2)^{1/2} \), apply the product and chain rules to find \( \frac{d^2y}{dx^2} \):\[ \frac{d}{dx}[-3x(2-x^2)^{1/2}] = -3 \left[(2-x^2)^{1/2} + x \cdot \frac{d}{dx}(2-x^2)^{1/2}\right] \]\[ \text{with } \frac{d}{dx}(2-x^2)^{1/2} = -\frac{x}{(2-x^2)^{1/2}} \]\[ = -3 \left[(2-x^2)^{1/2} - \frac{x^2}{(2-x^2)^{1/2}} \right] \]\[ = -3 \left( \frac{2-x^2-x^2}{(2-x^2)^{1/2}} \right) \]\[ = -3 \left( \frac{2 - 2x^2}{(2-x^2)^{1/2}} \right) \]\[ = -\frac{3(2 - 2x^2)}{(2-x^2)^{1/2}} \].

Step 4: Determine Inflection Points

Set the second derivative equal to zero to find inflection points:\[ -3 \left( \frac{2 - 2x^2}{(2-x^2)^{1/2}} \right) = 0 \]Simplifying yields:\[ 2 - 2x^2 = 0 \]Solving gives \(x^2 = 1\), or \(x = \pm 1\). So, the inflection points occur at \(x = 1 \text{ and } x = -1\).

Step 5: Classify Critical Points and Inflection Points

Evaluate the function value at each critical point: At \(x = 0\), \(y = (2-0)^{3/2} = 2^{3/2}\).At \(x = \sqrt{2} \text{ and } x = -\sqrt{2}\), \(y = (2 - 2)^{3/2} = 0\). For inflection points, evaluate:\( y(1) = (2-1^2)^{3/2} = 1 \)and \( y(-1) = (2-(-1)^2)^{3/2} = 1 \).Thus, local extrema are found at \((0, 2^{3/2})\) and the function intersects the x-axis at \((\pm \sqrt{2}, 0)\). The inflection points are \((1, 1)\) and \((-1, 1)\).

Step 6: Graph the Function

To graph the function, plot the critical points and inflection points found in earlier steps:- Local maximum at \((0, 2^{3/2})\).- X-intercepts at \((\pm \sqrt{2}, 0)\).- Inflection points at \((1, 1)\) and \((-1, 1)\).Draw a smooth curve passing through these points, showing the local maximum, and note the change in concavity at the inflection points.

Key Concepts

Extreme points

Extreme points, also known as maxima or minima, are points where a function reaches a peak (maximum) or a valley (minimum) in a given interval. For the function \( y = (2 - x^2)^{3/2} \), we look for extreme points by first determining its critical points, where the derivative is zero or undefined.

In the solution process for our exercise, we found the derivative \( \frac{dy}{dx} = -3x(2-x^2)^{1/2} \). Setting this equal to zero helps us find critical points: \( x = 0, \pm \sqrt{2} \). At these values, the behavior of the function changes drastically—either reaching a maximum height, a minimum depth, or a horizontal tangent line.

Ultimately, by calculating \( y(0) = 2^{3/2} \), we see that the function has a local maximum at \((0, 2^{3/2})\). Exploring these extremes is crucial for understanding the shape of the function's graph and can help us identify optimal solutions in various practical problems.

Inflection points

Inflection points are where the function's concavity changes. This means the shape of the graph shifts from concave up (holding water) to concave down (spilling water), or vice versa. To locate these points, we examine the second derivative of the function for where it changes sign.

For our function \( y = (2 - x^2)^{3/2} \), the second derivative is quite complex but has been simplified to: \(-\frac{3(2 - 2x^2)}{(2-x^2)^{1/2}}\). By setting the second derivative to zero, we find the potential inflection points: \( x = 1 \) and \( x = -1 \).

Calculating the corresponding function values shows that at \((1, 1)\) and \((-1, 1)\), the function shifts concavity. Understanding inflection points is crucial for visualizing how a function behaves over an interval, assisting in fields like physics for understanding motion or in economics for analyzing cost functions.

Chain rule

The Chain Rule is an essential tool in calculus for differentiating composite functions, where one function is nested inside another. It breaks down complex derivatives into manageable steps by introducing intermediate variables.

In the exercise, the chain rule was employed to differentiate \(y = (2 - x^2)^{3/2}\). By setting \(u = 2 - x^2\), the function becomes \(y = u^{3/2}\). The rule states that the derivative of a composite function is the derivative of the outside function evaluated at the inside function times the derivative of the inside function:

• The outside function: \(u^{3/2}\), derivative \(\frac{3}{2}u^{1/2}\)
• The inside function: \(2-x^2\), derivative \(-2x\)

Thus, combining these, \(\frac{dy}{dx} = \frac{3}{2} (2-x^2)^{1/2} (-2x) = -3x(2-x^2)^{1/2}\). This approach simplifies differentiation for complex functions, making it invaluable across various disciplines.

Critical points

Critical points are where the derivative of a function is zero or undefined, indicating potential local minima and maxima or points of inflection. These points are often where significant changes occur in the behavior of the function.

In the example \( y = (2 - x^2)^{3/2} \), the critical points are found by solving \(-3x(2-x^2)^{1/2} = 0\). This solution gives us \( x = 0, \pm \sqrt{2} \), where either the derivative equals zero or is undefined.

At these critical points, the function may reach a peak or trough, or simply level out temporarily. By evaluating function values at these points, it becomes possible to classify them as local minima, maxima, or just saddle points. Recognizing critical points is fundamental in many applied fields, from engineering to optimization problems.