Identify The Open Intervals On Which The Function Is Increasing Or Decreasing. (Enter Your Answers Using Interval Notation.)${ Y = X \sqrt{100-x^2} }$- Increasing: $\square$- Decreasing: $\square$

Introduction

In calculus, understanding the behavior of functions is crucial for various applications, including optimization problems and modeling real-world phenomena. One essential concept in this context is the determination of intervals where a function is increasing or decreasing. This article focuses on identifying the open intervals on which the given function, $y = x \sqrt{100-x^2}$, is increasing or decreasing.

Understanding the Function

The given function is a product of two functions: $x$ and $\sqrt{100-x^2}$. To analyze its behavior, we need to consider the properties of each component. The function $x$ is a linear function that increases as $x$ increases. On the other hand, the function $\sqrt{100-x^2}$ is a square root function that is defined only for $-10 \leq x \leq 10$. This function is increasing on the interval $[-10, 0)$ and decreasing on the interval $(0, 10]$.

Finding the Critical Points

To determine the intervals where the function is increasing or decreasing, we need to find the critical points. Critical points occur where the derivative of the function is equal to zero or undefined. In this case, we need to find the derivative of the function using the product rule.

Calculating the Derivative

Using the product rule, the derivative of the function is given by:

$$\frac{dy}{dx} = \sqrt{100-x^2} + x \cdot \frac{1}{2\sqrt{100-x^2}} \cdot (-2x)$$

Simplifying the expression, we get:

$$\frac{dy}{dx} = \sqrt{100-x^2} - \frac{x^2}{\sqrt{100-x^2}}$$

Simplifying the Derivative

We can simplify the derivative further by combining the two terms:

$$\frac{dy}{dx} = \frac{100-x^2-x^2}{\sqrt{100-x^2}}$$

$$\frac{dy}{dx} = \frac{100-2x^2}{\sqrt{100-x^2}}$$

Finding the Critical Points

To find the critical points, we need to set the derivative equal to zero and solve for $x$:

$$\frac{100-2x^2}{\sqrt{100-x^2}} = 0$$

Since the denominator is always positive, we can set the numerator equal to zero:

$$100-2x^2 = 0$$

Solving for $x$, we get:

$$x^2 = 50$$

$$x = \pm \sqrt{50}$$

$$x = \pm 5\sqrt{2}$$

Analyzing the Intervals

Now that we have found the critical points, we can analyze the intervals where the function is increasing or decreasing. We need to test the sign of the derivative in each interval.

Interval $(-10, -5\sqrt{2})$

In this interval, we can choose a test point, say $x = -6$. Plugging this value into the derivative, we get:

$$\frac{dy}{dx} = \frac{100-2(-6)^2}{\sqrt{100-(-6)^2}}$$

$$\frac{dy}{dx} = \frac{100-72}{\sqrt{100-36}}$$

$$\frac{dy}{dx} = \frac{28}{\sqrt{64}}$$

$$\frac{dy}{dx} = \frac{28}{8}$$

$$\frac{dy}{dx} = 3.5$$

Since the derivative is positive, the function is increasing in this interval.

Interval $(-5\sqrt{2}, 5\sqrt{2})$

In this interval, we can choose a test point, say $x = 0$. Plugging this value into the derivative, we get:

$$\frac{dy}{dx} = \frac{100-2(0)^2}{\sqrt{100-(0)^2}}$$

$$\frac{dy}{dx} = \frac{100}{\sqrt{100}}$$

$$\frac{dy}{dx} = \frac{100}{10}$$

$$\frac{dy}{dx} = 10$$

Since the derivative is positive, the function is increasing in this interval.

Interval $(5\sqrt{2}, 10)$

In this interval, we can choose a test point, say $x = 6$. Plugging this value into the derivative, we get:

$$\frac{dy}{dx} = \frac{100-2(6)^2}{\sqrt{100-(6)^2}}$$

$$\frac{dy}{dx} = \frac{100-72}{\sqrt{100-36}}$$

$$\frac{dy}{dx} = \frac{28}{\sqrt{64}}$$

$$\frac{dy}{dx} = \frac{28}{8}$$

$$\frac{dy}{dx} = 3.5$$

Since the derivative is positive, the function is increasing in this interval.

Conclusion

In conclusion, the function $y = x \sqrt{100-x^2}$ is increasing on the intervals $(-10, -5\sqrt{2})$, $(-5\sqrt{2}, 5\sqrt{2})$, and $(5\sqrt{2}, 10)$. The function is decreasing on the interval $(-5\sqrt{2}, 5\sqrt{2})$.

Final Answer

• Increasing: $(-10, -5\sqrt{2}) \cup (-5\sqrt{2}, 5\sqrt{2}) \cup (5\sqrt{2}, 10)$
• Decreasing: $(-5\sqrt{2}, 5\sqrt{2})$
  

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Introduction

In our previous article, we analyzed the behavior of the function $y = x \sqrt{100-x^2}$ and determined the intervals where the function is increasing or decreasing. In this article, we will answer some frequently asked questions related to this topic.

Q1: What is the domain of the function $y = x \sqrt{100-x^2}$?

A1: The domain of the function is $-10 \leq x \leq 10$, since the expression under the square root must be non-negative.

Q2: How do you find the critical points of the function?

A2: To find the critical points, we need to set the derivative of the function equal to zero and solve for $x$. In this case, we found the critical points to be $x = \pm 5\sqrt{2}$.

Q3: What is the significance of the critical points?

A3: The critical points divide the domain of the function into intervals where the function is increasing or decreasing. By analyzing the sign of the derivative in each interval, we can determine the behavior of the function.

Q4: How do you determine the sign of the derivative in each interval?

A4: To determine the sign of the derivative, we can choose a test point in each interval and plug it into the derivative. If the derivative is positive, the function is increasing in that interval. If the derivative is negative, the function is decreasing in that interval.

Q5: What is the relationship between the derivative and the behavior of the function?

A5: The derivative of the function represents the rate of change of the function with respect to $x$. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing.

Q6: Can you provide an example of how to use the derivative to determine the behavior of the function?

A6: Let's say we want to determine the behavior of the function in the interval $(-10, -5\sqrt{2})$. We can choose a test point, say $x = -6$, and plug it into the derivative. If the derivative is positive, the function is increasing in that interval.

Q7: How do you use the derivative to determine the intervals where the function is increasing or decreasing?

A7: To determine the intervals where the function is increasing or decreasing, we need to analyze the sign of the derivative in each interval. We can do this by choosing a test point in each interval and plugging it into the derivative.

Q8: What is the significance of the intervals where the function is increasing or decreasing?

A8: The intervals where the function is increasing or decreasing are important because they help us understand the behavior of the function. By knowing where the function is increasing or decreasing, we can make informed decisions about the function's behavior.

Q9: Can you provide an example of how to use the intervals where the function is increasing or decreasing to make informed decisions?

A9: Let's say we want to maximize the function. We can do this by finding the interval where the function is increasing and then choosing a point in that interval that maximizes the function.

Q10: How do you use the intervals where the function is increasing or decreasing to maximize or minimize the function?

A10: To maximize or minimize the function, we need to analyze the intervals where the function is increasing or decreasing and then choose a point in that interval that maximizes or minimizes the function.

Conclusion

In conclusion, understanding the behavior of the function $y = x \sqrt{100-x^2}$ is crucial for various applications, including optimization problems and modeling real-world phenomena. By analyzing the intervals where the function is increasing or decreasing, we can make informed decisions about the function's behavior and use it to maximize or minimize the function.

Final Answer

• Increasing: $(-10, -5\sqrt{2}) \cup (-5\sqrt{2}, 5\sqrt{2}) \cup (5\sqrt{2}, 10)$
• Decreasing: $(-5\sqrt{2}, 5\sqrt{2})$