Problem 33 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$g(x)=x \sqrt{8-x^{2}}$$

Step-by-Step Solution

Answer

Function is increasing on \((-\sqrt{8}, -2) \cup (2, \sqrt{8})\) and decreasing on \((-2, 2)\). Local min at \(-2\) is \(-4\); local max at \(2\) is \(4\).

Step 1: Find the Domain

To find where the function \(g(x) = x \sqrt{8 - x^2}\) is defined, set the expression inside the square root greater than or equal to zero: \(8 - x^2 \geq 0\). This gives \(-\sqrt{8} \leq x \leq \sqrt{8}\). Thus, the domain is \([-\sqrt{8}, \sqrt{8}]\).

Step 2: Find the Derivative

Differentiate \(g(x) = x \sqrt{8 - x^2}\) using the product rule: \(g'(x) = \frac{d}{dx}[x] \times \sqrt{8-x^2} + x \times \frac{d}{dx}[\sqrt{8-x^2}]\). Calculate: \(g'(x) = \sqrt{8-x^2} + x \left(\frac{-x}{\sqrt{8-x^2}}\right) = \frac{8-2x^2}{\sqrt{8-x^2}}\).

Step 3: Determine Critical Points

Set the derivative \(g'(x)\) equal to zero to find critical points: \(\frac{8 - 2x^2}{\sqrt{8-x^2}} = 0\). This simplifies to \(8 - 2x^2 = 0\), giving \(x^2 = 4\) or \(x = \pm 2\). Critical points occur at \(x = 2\) and \(x = -2\).

Step 4: Determine Intervals of Increase and Decrease

Use a number line and test points in each interval created by critical points \(x = -2\), \(2\). Test \(x = -3\), \(x = 0\), and \(x = 3\) within their intervals:- For \(-\sqrt{8} < x < -2\), \(g'(x) > 0\), \(g(x)\) is increasing.- For \(-2 < x < 2\), \(g'(x) < 0\), \(g(x)\) is decreasing.- For \(2 < x < \sqrt{8}\), \(g'(x) > 0\), \(g(x)\) is increasing.

Step 5: Identify Local and Absolute Extreme Values

Evaluate \(g(x)\) at critical points and endpoints: - \(g(-\sqrt{8}) = 0\), \(g(-2) = -4\), \(g(2) = 4\), \(g(\sqrt{8}) = 0\).- \(x = -2\) gives a local minimum \(-4\);- \(x = 2\) gives a local maximum \(4\);No values outside critical points outperform maximum/minimum.

Key Concepts

Critical Points

Understanding critical points is essential for analyzing a function's behavior. Critical points occur where the derivative of a function is zero or undefined. These points help identify where the function might change direction or have a peak or valley.
In our exercise, we determined the critical points by first finding the derivative of the function, which led to the equation \(8 - 2x^2 = 0\). Solving this gives \(x = \pm 2\), so the critical points are \(x = 2\) and \(x = -2\).
These points are crucial because they help separate the function into intervals, where it could either be increasing or decreasing. Knowing where these changes occur can provide insights into the local extreme values and the overall shape of the graph.

Local Extreme Values

Local extreme values describe the points on a graph where the function reaches a local minimum or maximum. These are the highest or lowest points within a certain interval, not necessarily over the entire range of the function.
To find these values in the exercise, we evaluated the function at the critical points and also at the endpoints of the domain. The critical points we found were \(x = -2\) and \(x = 2\), with their corresponding function values calculated as \(-4\) and \(4\) respectively. This showed that the function had a local minimum of \(-4\) at \(x = -2\) and a local maximum of \(4\) at \(x = 2\).
By examining these values, we can understand where the function takes its highest or lowest turn locally, providing invaluable insights into its graph's structure and behavior.

Derivative Test

The derivative test is a powerful tool for finding whether a function is increasing or decreasing, as well as for identifying local extremes. By examining the sign of the derivative in different intervals, we obtain a clearer picture of the function's dynamics.
In practice, you compute the derivative of the function and check its value at selected points between and around critical points. In our exercise, we looked at test points in the intervals \((-\sqrt{8}, -2)\), \((-2, 2)\), and \((2, \sqrt{8})\). Based on the derivative's value, we classified these intervals as follows:

• \(-\sqrt{8} < x < -2\): the derivative is positive \( (g'(x) > 0)\), so the function is increasing.
• \(-2 < x < 2\): the derivative is negative \( (g'(x) < 0)\), so the function is decreasing.
• \(2 < x < \sqrt{8}\): the derivative is positive \( (g'(x) > 0)\), so the function is increasing.

This analysis reveals where the function ascends or descends and helps identify potential local maxima and minima, sharpening our comprehension of the function's shape.