Predicting True Statements

Oct 13, 2025 by magicalceylontours.com 57 views

Understanding the Problem

The given table represents a continuous function $f(x)$ with specific values at certain points. We are asked to predict which statements about the intervals of this function are true. To approach this problem, we need to analyze the behavior of the function between the given points and make informed decisions about the intervals.

Analyzing the Function

From the table, we can see that the function has the following values:

$x$	$f(x)$
-3	-15
-2	0
-1	3
0	0

We can observe that the function changes from a negative value to a positive value between $x = -3$ and $x = -2$ . This indicates that the function has a local minimum at $x = -2$ . Similarly, the function changes from a positive value to a negative value between $x = 0$ and $x = -1$ . This indicates that the function has a local maximum at $x = 0$ .

Predicting True Statements

Based on the analysis of the function, we can make the following predictions about the intervals:

Statement 1: The function is decreasing on the interval $(-3, -2)$

The function has a negative value at $x = -3$ and a positive value at $x = -2$ . Since the function is continuous, it must be decreasing on the interval $(-3, -2)$ .

Statement 2: The function is increasing on the interval $(-2, -1)$

The function has a positive value at $x = -2$ and a positive value at $x = -1$ . Since the function is continuous, it must be increasing on the interval $(-2, -1)$ .

Statement 3: The function is decreasing on the interval $(-1, 0)$

The function has a positive value at $x = -1$ and a negative value at $x = 0$ . Since the function is continuous, it must be decreasing on the interval $(-1, 0)$ .

Statement 4: The function has a local minimum at $x = -2$

As mentioned earlier, the function changes from a negative value to a positive value between $x = -3$ and $x = -2$ . This indicates that the function has a local minimum at $x = -2$ .

Statement 5: The function has a local maximum at $x = 0$

As mentioned earlier, the function changes from a positive value to a negative value between $x = 0$ and $x = -1$ . This indicates that the function has a local maximum at $x = 0$ .

Statement 6: The function is continuous on the interval $(-3, 0)$

Since the function has no gaps or jumps on the interval $(-3, 0)$ , it must be continuous on this interval.

Statement 7: The function is differentiable on the interval $(-3, 0)$

Since the function is continuous on the interval $(-3, 0)$ , it must also be differentiable on this interval.

Statement 8: The function has a derivative of 0 at $x = -2$

Since the function is increasing on the interval $(-2, -1)$ , its derivative must be positive. However, since the function is decreasing on the interval $(-1, 0)$ , its derivative must be negative. Therefore, the derivative of the function at $x = -2$ must be 0.

Statement 9: The function has a second derivative of 0 at $x = -2$

Since the derivative of the function at $x = -2$ is 0, its second derivative must also be 0.

Statement 10: The function has a local minimum at $x = -1$

Since the function is decreasing on the interval $(-1, 0)$ , it must have a local minimum at $x = -1$ .

Statement 11: The function has a local maximum at $x = -3$

Since the function is decreasing on the interval $(-3, -2)$ , it must have a local maximum at $x = -3$ .

Statement 12: The function is continuous on the interval $(-3, -1)$

Since the function has no gaps or jumps on the interval $(-3, -1)$ , it must be continuous on this interval.

Statement 13: The function is differentiable on the interval $(-3, -1)$

Since the function is continuous on the interval $(-3, -1)$ , it must also be differentiable on this interval.

Statement 14: The function has a derivative of 0 at $x = -1$

Since the function is decreasing on the interval $(-1, 0)$ , its derivative must be negative. However, since the function is increasing on the interval $(-2, -1)$ , its derivative must be positive. Therefore, the derivative of the function at $x = -1$ must be 0.

Statement 15: The function has a second derivative of 0 at $x = -1$

Since the derivative of the function at $x = -1$ is 0, its second derivative must also be 0.

Statement 16: The function has a local minimum at $x = 0$

Since the function is decreasing on the interval $(-1, 0)$ , it must have a local minimum at $x = 0$ .

Statement 17: The function has a local maximum at $x = -2$

Since the function is increasing on the interval $(-2, -1)$ , it must have a local maximum at $x = -2$ .

Statement 18: The function is continuous on the interval $(-2, 0)$

Since the function has no gaps or jumps on the interval $(-2, 0)$ , it must be continuous on this interval.

Statement 19: The function is differentiable on the interval $(-2, 0)$

Since the function is continuous on the interval $(-2, 0)$ , it must also be differentiable on this interval.

Statement 20: The function has a derivative of 0 at $x = 0$

Statement 21: The function has a second derivative of 0 at $x = 0$

Since the derivative of the function at $x = 0$ is 0, its second derivative must also be 0.

Statement 22: The function has a local minimum at $x = -3$

Since the function is decreasing on the interval $(-3, -2)$ , it must have a local minimum at $x = -3$ .

Statement 23: The function has a local maximum at $x = 0$

Since the function is decreasing on the interval $(-1, 0)$ , it must have a local maximum at $x = 0$ .

Statement 24: The function is continuous on the interval $(-3, -2)$

Since the function has no gaps or jumps on the interval $(-3, -2)$ , it must be continuous on this interval.

Statement 25: The function is differentiable on the interval $(-3, -2)$

Since the function is continuous on the interval $(-3, -2)$ , it must also be differentiable on this interval.

Statement 26: The function has a derivative of 0 at $x = -3$

Since the function is decreasing on the interval $(-3, -2)$ , its derivative must be negative. However, since the function is increasing on the interval $(-2, -1)$ , its derivative must be positive. Therefore, the derivative of the function at $x = -3$ must be 0.

Statement 27: The function has a second derivative of 0 at $x = -3$

Since the derivative of the function at $x = -3$ is 0, its second derivative must also be 0.

Statement 28: The function has a local minimum at $x = 0$

Since the function is decreasing on the interval $(-1, 0)$ , it must have a local minimum at $x = 0$ .

Statement 29: The function has a local maximum at $x = -2$

Since the function is increasing on the interval $(-2, -1)$ , it must have a local maximum at $x = -2$ .

Statement 30: The function is continuous on the interval $(-2, -1)$

Since the function has no gaps or jumps on the interval $(-2, -1)$ , it must be continuous on this interval.

Statement 31: The function is differentiable on the interval $(-2, -1)$

Since the function is continuous on the interval $(-2, -1)$ , it must also be differentiable on this interval.

Statement 32: The function has a derivative of 0 at $x = -2$

Q: What is a continuous function?

A: A continuous function is a function that has no gaps or jumps in its graph. It can be thought of as a function that can be drawn without lifting the pencil from the paper.

Q: What is the significance of a continuous function?

A: A continuous function is significant because it can be used to model real-world phenomena, such as the motion of an object or the growth of a population. It is also used in many mathematical and scientific applications, such as calculus and physics.

Q: How do I determine if a function is continuous?

A: To determine if a function is continuous, you need to check if it has any gaps or jumps in its graph. You can do this by plotting the function on a graph and looking for any breaks or discontinuities.

Q: What is the difference between a local minimum and a local maximum?

A: A local minimum is a point on the graph of a function where the function is decreasing and then increases. A local maximum is a point on the graph of a function where the function is increasing and then decreases.

Q: How do I determine if a function has a local minimum or a local maximum?

A: To determine if a function has a local minimum or a local maximum, you need to look at the behavior of the function around the point in question. If the function is decreasing and then increases, it has a local minimum. If the function is increasing and then decreases, it has a local maximum.

Q: What is the significance of a local minimum and a local maximum?

A: A local minimum and a local maximum are significant because they can be used to determine the behavior of a function over a given interval. They can also be used to find the maximum or minimum value of a function.

Q: How do I determine if a function is differentiable?

A: To determine if a function is differentiable, you need to check if it has any sharp corners or cusps in its graph. You can do this by plotting the function on a graph and looking for any sharp corners or cusps.

Q: What is the difference between a derivative and a second derivative?

A: A derivative is a measure of how fast a function is changing at a given point. A second derivative is a measure of how fast the derivative is changing at a given point.

Q: How do I determine if a function has a derivative of 0?

A: To determine if a function has a derivative of 0, you need to look at the behavior of the function around the point in question. If the function is increasing and then decreases, its derivative is 0. If the function is decreasing and then increases, its derivative is 0.

Q: What is the significance of a derivative of 0?

A: A derivative of 0 is significant because it can be used to determine the behavior of a function over a given interval. It can also be used to find the maximum or minimum value of a function.

Q: How do I determine if a function has a second derivative of 0?

A: To determine if a function has a second derivative of 0, you need to look at the behavior of the function around the point in question. If the derivative is 0, its second derivative is also 0.

Q: What is the significance of a second derivative of 0?

A: A second derivative of 0 is significant because it can be used to determine the behavior of a function over a given interval. It can also be used to find the maximum or minimum value of a function.

Q: How do I use the statements to solve problems?

A: To use the statements to solve problems, you need to apply the concepts of continuity, differentiability, and local minima and maxima to the given function. You can use the statements to determine the behavior of the function over a given interval and to find the maximum or minimum value of the function.

Q: What are some common applications of the statements?

A: Some common applications of the statements include:

Modeling real-world phenomena, such as the motion of an object or the growth of a population
Calculus and physics
Optimization problems, such as finding the maximum or minimum value of a function
Economics and finance, such as modeling the behavior of stock prices or interest rates

Q: How do I practice using the statements?

A: To practice using the statements, you can try solving problems that involve continuity, differentiability, and local minima and maxima. You can also try applying the statements to real-world problems, such as modeling the behavior of a population or the growth of a company.