ListenWhich Statements Are True About The Function $f(x)=x^3-x^2-4x+4$? Select All That Apply.A. The Function Is Positive Over The Intervals $(-2,1$\] And $(2, \infty$\], And The Function Is Negative Over The Interval

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The given function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 is a cubic function, which means it can have up to three real roots. To understand the behavior of this function, we need to analyze its graph and identify the intervals where it is positive or negative.

Finding the Critical Points

To find the critical points of the function, we need to find the values of xx where the derivative of the function is equal to zero or undefined. The derivative of the function f(x)f(x) is given by:

f′(x)=3x2−2x−4f'(x) = 3x^2 - 2x - 4

To find the critical points, we need to solve the equation f′(x)=0f'(x) = 0. We can do this by factoring the quadratic expression:

3x2−2x−4=(3x+2)(x−2)=03x^2 - 2x - 4 = (3x + 2)(x - 2) = 0

This gives us two possible values for xx: x=−23x = -\frac{2}{3} and x=2x = 2. These are the critical points of the function.

Analyzing the Intervals

To analyze the intervals where the function is positive or negative, we need to examine the behavior of the function in each interval. We can do this by choosing a test point in each interval and evaluating the function at that point.

Interval (−∞,−23)(-\infty, -\frac{2}{3})

Let's choose the test point x=−1x = -1. Evaluating the function at this point, we get:

f(−1)=(−1)3−(−1)2−4(−1)+4=−1−1+4+4=6f(-1) = (-1)^3 - (-1)^2 - 4(-1) + 4 = -1 - 1 + 4 + 4 = 6

Since the function is positive at this point, we can conclude that the function is positive over the interval (−∞,−23)(-\infty, -\frac{2}{3}).

Interval (−23,2)(-\frac{2}{3}, 2)

Let's choose the test point x=0x = 0. Evaluating the function at this point, we get:

f(0)=(0)3−(0)2−4(0)+4=4f(0) = (0)^3 - (0)^2 - 4(0) + 4 = 4

Since the function is positive at this point, we can conclude that the function is positive over the interval (−23,2)(-\frac{2}{3}, 2).

Interval (2,∞)(2, \infty)

Let's choose the test point x=3x = 3. Evaluating the function at this point, we get:

f(3)=(3)3−(3)2−4(3)+4=27−9−12+4=10f(3) = (3)^3 - (3)^2 - 4(3) + 4 = 27 - 9 - 12 + 4 = 10

Since the function is positive at this point, we can conclude that the function is positive over the interval (2,∞)(2, \infty).

Conclusion

Based on our analysis, we can conclude that the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 is positive over the intervals (−∞,−23)(-\infty, -\frac{2}{3}) and (2,∞)(2, \infty), and the function is negative over the interval (−23,2)(-\frac{2}{3}, 2).

True Statements

  • The function is positive over the intervals (−∞,−23)(-\infty, -\frac{2}{3}) and (2,∞)(2, \infty).
  • The function is negative over the interval (−23,2)(-\frac{2}{3}, 2).

False Statements

  • The function is positive over the interval (1,2)(1, 2).
  • The function is negative over the interval (−2,1)(-2, 1).

Additional Analysis

To further analyze the function, we can examine its behavior at the critical points. We can do this by evaluating the function at the critical points and examining the behavior of the function in the vicinity of these points.

At the critical point x=−23x = -\frac{2}{3}, we have:

f(−23)=(−23)3−(−23)2−4(−23)+4=−827−49+83+4=8027f(-\frac{2}{3}) = (-\frac{2}{3})^3 - (-\frac{2}{3})^2 - 4(-\frac{2}{3}) + 4 = -\frac{8}{27} - \frac{4}{9} + \frac{8}{3} + 4 = \frac{80}{27}

Since the function is positive at this point, we can conclude that the function is increasing in the vicinity of this point.

At the critical point x=2x = 2, we have:

f(2)=(2)3−(2)2−4(2)+4=8−4−8+4=0f(2) = (2)^3 - (2)^2 - 4(2) + 4 = 8 - 4 - 8 + 4 = 0

Since the function is zero at this point, we can conclude that the function has a root at this point.

Conclusion

In our previous article, we analyzed the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 and concluded that it is positive over the intervals (−∞,−23)(-\infty, -\frac{2}{3}) and (2,∞)(2, \infty), and the function is negative over the interval (−23,2)(-\frac{2}{3}, 2). In this article, we will answer some frequently asked questions about this function.

Q: What is the domain of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4?

A: The domain of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 is all real numbers, since the function is defined for all values of xx.

Q: What is the range of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4?

A: The range of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 is all real numbers, since the function can take on any value.

Q: What are the critical points of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4?

A: The critical points of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 are x=−23x = -\frac{2}{3} and x=2x = 2.

Q: What is the behavior of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 in the vicinity of the critical points?

A: In the vicinity of the critical point x=−23x = -\frac{2}{3}, the function is increasing. In the vicinity of the critical point x=2x = 2, the function has a root.

Q: What is the significance of the critical points of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4?

A: The critical points of the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 are significant because they determine the behavior of the function in the vicinity of these points. The critical points also help us understand the shape of the graph of the function.

Q: How can we use the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 in real-world applications?

A: The function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 can be used in various real-world applications, such as modeling population growth, chemical reactions, and electrical circuits.

Q: Can we use the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 to solve optimization problems?

A: Yes, we can use the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 to solve optimization problems. For example, we can use the function to find the maximum or minimum value of a function.

Q: How can we graph the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4?

A: We can graph the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 by using a graphing calculator or a computer algebra system. We can also use the function to create a table of values and plot the points on a coordinate plane.

Q: Can we use the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 to model real-world phenomena?

A: Yes, we can use the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4 to model real-world phenomena, such as population growth, chemical reactions, and electrical circuits.

Conclusion

In this article, we answered some frequently asked questions about the function f(x)=x3−x2−4x+4f(x)=x^3-x^2-4x+4. We also discussed the significance of the critical points of the function and how it can be used in real-world applications.