Problem 37 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ g(x)=\sqrt{16-x^{2}} $$

Step-by-Step Solution

Answer

g(x) is increasing on (0, 4) and decreasing on (-4, 0).

Step 1: Understand the function domain

The function is defined by the square root \(\sqrt{16-x^2}\). To ensure the expression under the square root is non-negative, we solve \(16-x^2 \geq 0\). This implies \( -4 \leq x \leq 4\). Thus, the domain of \(g(x)\) is \([-4, 4]\).

Step 2: Find the derivative

To determine where the function is increasing or decreasing, find its derivative. The function can be rewritten as \(g(x) = (16-x^2)^{1/2}\). Using the chain rule, \(abla g(x) = \frac{d}{dx}(16-x^2)^{1/2} = \frac{1}{2}(16-x^2)^{-1/2}(-2x) = -\frac{x}{\sqrt{16-x^2}}\).

Step 3: Determine critical points

Set the derivative equal to zero to find the critical points: \(-\frac{x}{\sqrt{16-x^2}} = 0\). Solving this gives \(x = 0\).

Step 4: Analyze the sign of the derivative

Examine the sign of the derivative \(-\frac{x}{\sqrt{16-x^2}}\) in the intervals determined by the critical point \(x = 0\): \([-4, 0)\) and \(0, 4]\).- For \(x \in (-4, 0)\), \(-\frac{x}{\sqrt{16-x^2}} > 0\), so the function is decreasing.- For \(x \in (0, 4)\), \(-\frac{x}{\sqrt{16-x^2}} < 0\), so the function is increasing.

Key Concepts

Critical Points

Critical points in calculus are where the derivative of a function is zero or undefined. These points are essential for understanding the behavior of a function, like where it reaches its maximum or minimum values. Given the function in the exercise, we computed the derivative. Then, we set it to zero to determine the critical points. In this case, the critical point is at \(x = 0\). Identifying this point helps us see where the function might change from decreasing to increasing or vice versa. Critical points are crucial because they often signal where important changes in the function's behavior occur, especially on a closed interval. To find them:

• Compute the derivative of the function.
• Set the derivative equal to zero.
• Solve for \(x\) to find the critical points.

Increasing and Decreasing Functions

Understanding when a function is increasing or decreasing helps us grasp its overall shape and behavior. A function is increasing on an interval if, as you move from left to right along the interval, the function goes upward. It's decreasing if it trends downward. In the given problem, the derivative \(-\frac{x}{\sqrt{16-x^2}}\) helps identify these intervals.If the derivative is positive over an interval, the function is increasing there. Conversely, if it's negative, the function is decreasing. For \([-4, 0)\), the derivative is positive, indicating the function decreases. For \((0, 4]\), it is negative, so the function increases. By analyzing the sign of the derivative, students can understand when and where a function's behavior changes.

Derivative

The derivative is a fundamental concept in calculus. It measures how a function's output value changes as its input changes. In simpler terms, it gives the rate of change of the function. It can describe how steep a slope is at any point of a curve. For example, the derivative of \(g(x)=\sqrt{16-x^2}\) is \(-\frac{x}{\sqrt{16-x^2}}\). This expression tells us how the function is changing at every point within its domain.Calculating derivatives involve rules such as the power rule, the product rule, and crucially, the chain rule, which was used here. The chain rule is especially helpful when dealing with composite functions, like \((16-x^2)^{1/2}\). Understanding derivatives is key to analyzing the behavior of functions, allowing us to find critical points, intervals of increase or decrease, and even concavity and points of inflection later on.