Range Of F(x) = (x+2)(x-2)^2 Identifying Impossible Values For Q

JU07/28/2025 08, 2025 by THE IDEN 65 views

Understanding the Range of a Cubic Function A Detailed Analysis of f(x) = (x+2)(x-2)^2

In this article, we delve into the intricacies of the function f(x) = (x+2)(x-2)^2, a cubic polynomial that presents an interesting challenge in determining its range within a specific domain. We will explore how to find the possible values of q for points (p, q) on the graph of y = f(x), where -2 ≤ p ≤ 2. Our focus will be on identifying which of the given options (12, 8, 4, 0) is NOT a possible value of q. This involves a combination of algebraic manipulation, calculus-based optimization techniques, and graphical analysis to provide a comprehensive understanding of the function's behavior.

Problem Statement: Exploring the Function f(x) = (x+2)(x-2)^2

The core of our discussion revolves around the function f(x) = (x+2)(x-2)^2. This cubic function is a product of a linear term (x+2) and a squared linear term (x-2)^2. The squared term indicates a repeated root at x = 2, which will have significant implications for the shape of the graph. The problem stipulates that we are interested in the portion of the graph within the vertical strip defined by -2 ≤ x ≤ 2. We are given a point (p, q) on the graph, where p falls within this interval, and our task is to determine which of the provided q values (12, 8, 4, 0) cannot be the function's output for any p in the given range. To achieve this, we must thoroughly analyze the function’s behavior, including its critical points, local extrema, and the values it attains at the boundaries of the interval.

Expanding and Analyzing the Cubic Function f(x)

To begin our analysis of the function f(x) = (x+2)(x-2)^2, let's first expand it to a more standard polynomial form. This will make it easier to apply calculus techniques for finding critical points and analyzing the function’s behavior. Expanding the expression, we get:

f(x) = (x+2)(x^2 - 4x + 4) f(x) = x^3 - 4x^2 + 4x + 2x^2 - 8x + 8 f(x) = x^3 - 2x^2 - 4x + 8

Now that we have the function in the form f(x) = x^3 - 2x^2 - 4x + 8, we can proceed with finding its derivative. The derivative, f'(x), will help us identify the critical points, which are the points where the function's slope is either zero or undefined. These points are crucial for determining the local maxima and minima of the function, which play a key role in finding the range of the function within the given interval. Taking the derivative:

f'(x) = 3x^2 - 4x - 4

To find the critical points, we set f'(x) = 0 and solve for x:

3x^2 - 4x - 4 = 0

This is a quadratic equation, and we can solve it either by factoring or using the quadratic formula. Factoring the quadratic expression, we look for two numbers that multiply to (3)(-4) = -12 and add up to -4. These numbers are -6 and 2. We can rewrite the middle term using these numbers and factor by grouping:

3x^2 - 6x + 2x - 4 = 0 3x(x - 2) + 2(x - 2) = 0 (3x + 2)(x - 2) = 0

Setting each factor equal to zero gives us the critical points:

3x + 2 = 0 => x = -2/3 x - 2 = 0 => x = 2

Thus, the critical points of the function are x = -2/3 and x = 2. These points, along with the endpoints of the interval (x = -2 and x = 2), are the candidates for local maxima and minima within the interval -2 ≤ x ≤ 2. Understanding these critical points is crucial for determining the possible values of q that the function can take within the specified domain.

Evaluating the Function at Critical Points and Endpoints

Having identified the critical points x = -2/3 and x = 2, as well as the endpoints of the interval x = -2 and x = 2, our next step is to evaluate the function f(x) = x^3 - 2x^2 - 4x + 8 at these points. This will give us the y-coordinates corresponding to these x-values, which are crucial for determining the function's range within the interval -2 ≤ x ≤ 2. By calculating the function values at these key points, we can identify potential local maxima, local minima, and the overall range of the function in the specified domain.

First, let’s evaluate the function at the left endpoint, x = -2:

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

Next, we evaluate the function at the critical point x = -2/3:

f(-2/3) = (-2/3)^3 - 2(-2/3)^2 - 4(-2/3) + 8 f(-2/3) = -8/27 - 2(4/9) + 8/3 + 8 f(-2/3) = -8/27 - 8/9 + 8/3 + 8 f(-2/3) = -8/27 - 24/27 + 72/27 + 216/27 f(-2/3) = 256/27 ≈ 9.48

Now, we evaluate the function at the other critical point, x = 2:

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

Finally, we evaluate the function at the right endpoint, x = 2. However, we have already calculated f(2) = 0, so we don't need to repeat the calculation.

From these evaluations, we have the following points:

(-2, 0)
(-2/3, 256/27)
(2, 0)

These points provide valuable information about the function's behavior within the interval -2 ≤ x ≤ 2. We see that the function has roots at x = -2 and x = 2, and a local maximum at x = -2/3. The maximum value of the function in this interval is approximately 9.48. This information is crucial for determining the possible values of q and identifying which of the given options is not a possible value.

Determining the Range and Identifying the Impossible Value of q

Based on our evaluations, we have a clearer picture of the function f(x) = x^3 - 2x^2 - 4x + 8 within the interval -2 ≤ x ≤ 2. We found the following points:

f(-2) = 0
f(-2/3) = 256/27 ≈ 9.48
f(2) = 0

These points reveal that the function has roots at x = -2 and x = 2, and a local maximum at x = -2/3, where the function value is approximately 9.48. Since the function is continuous within this interval, it will take on all values between its minimum and maximum. The minimum value within the interval is 0, which occurs at both endpoints and the critical point x = 2. The maximum value is approximately 9.48, which occurs at the critical point x = -2/3.

Therefore, the range of the function within the interval -2 ≤ x ≤ 2 is 0 ≤ q ≤ 256/27 (approximately 0 to 9.48). Now, we can compare this range to the given options for q: A) 12, B) 8, C) 4, and D) 0.

A) 12: This value is outside the range 0 ≤ q ≤ 256/27, as 12 is greater than 9.48.
B) 8: This value falls within the range, as 0 ≤ 8 ≤ 9.48.
C) 4: This value also falls within the range, as 0 ≤ 4 ≤ 9.48.
D) 0: This value is the minimum value of the function in the interval and is therefore within the range.

From this comparison, it is clear that the value q = 12 is NOT a possible value for the function within the given interval. The function's maximum value is approximately 9.48, and therefore it cannot reach a value of 12.

Conclusion: Identifying the Impossible Value

In conclusion, by expanding and analyzing the cubic function f(x) = (x+2)(x-2)^2, finding its critical points, evaluating the function at these points and the endpoints of the interval -2 ≤ x ≤ 2, we were able to determine the range of the function within the specified domain. The range was found to be 0 ≤ q ≤ 256/27, which is approximately 0 to 9.48.

Comparing this range to the given options, we identified that q = 12 is the only value that falls outside this range. Therefore, 12 is NOT a possible value of q for any point (p, q) on the graph of y = f(x) where -2 ≤ p ≤ 2. This comprehensive analysis demonstrates how a combination of algebraic manipulation and calculus techniques can be used to understand the behavior of a function and determine its range within a given interval.