Critical Points And Local Minimum Values For F(x) = 12x^4 - 6x^3

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Introduction to Critical Points

In calculus, identifying critical points is a fundamental step in understanding the behavior of a function. Critical points are the points where the derivative of the function is either zero or undefined. These points are crucial because they often indicate where a function reaches its local maxima, local minima, or saddle points. Understanding critical points allows us to analyze the function's increasing and decreasing intervals, which is essential for sketching its graph and solving optimization problems.

For a given function f(x), critical points are found by first calculating its derivative, f'(x). We then solve for the values of x where f'(x) = 0 or where f'(x) is undefined. These x-values are the critical points of the function. It's important to consider the domain of the function when identifying critical points; only those points within the function's domain are relevant. For instance, if we are analyzing a function over a closed interval, we must also consider the endpoints of the interval as potential locations for extrema. The critical points, along with the endpoints, are the candidates for local and global maxima and minima.

Critical points play a vital role in various applications across different fields, such as physics, economics, and engineering. In physics, they can represent equilibrium points in a system. In economics, they can help determine the optimal production level or price to maximize profit. In engineering, they are used to design structures that can withstand maximum stress or to optimize the performance of a system. Therefore, mastering the technique of finding and analyzing critical points is crucial for anyone working with mathematical models in real-world applications. By examining the critical points, we gain valuable insights into the function's behavior, enabling us to make informed decisions and predictions based on the function's properties. This process not only helps in understanding the local behavior but also in determining the global behavior of the function over its entire domain.

Finding Critical Points for f(x) = 12x^4 - 6x^3

To find the critical points for the function f(x) = 12x^4 - 6x^3 on the interval [-1, 1], we need to follow a systematic approach. The first step involves calculating the derivative of the function. The derivative, denoted as f'(x), represents the instantaneous rate of change of the function at any point x. Using the power rule, we differentiate f(x) as follows:

f(x) = 12x^4 - 6x^3

Applying the power rule, which states that the derivative of x^n is nx^(n-1), we get:

f'(x) = 12(4x^3) - 6(3x^2)

Simplifying this expression, we obtain:

f'(x) = 48x^3 - 18x^2

Now that we have the derivative, the next step is to find the values of x for which f'(x) = 0 or where f'(x) is undefined. In this case, f'(x) is a polynomial, so it is defined for all real numbers. Therefore, we only need to solve for the values of x where f'(x) = 0.

Setting f'(x) = 0, we have:

48x^3 - 18x^2 = 0

To solve this equation, we can factor out the common terms. Both terms have x^2 as a factor, and the greatest common divisor of 48 and 18 is 6, so we can factor out 6x^2:

6x^2(8x - 3) = 0

This equation is satisfied if either 6x^2 = 0 or (8x - 3) = 0. Solving 6x^2 = 0, we get:

x^2 = 0

x = 0

Solving (8x - 3) = 0, we get:

8x = 3

x = 3/8

Thus, the critical points of the function f(x) = 12x^4 - 6x^3 are x = 0 and x = 3/8. These are the points where the slope of the tangent line to the curve is zero, indicating potential local maxima or minima. Additionally, since we are considering the interval [-1, 1], we also need to include the endpoints of the interval as potential critical points. Therefore, the critical points in the interval [-1, 1] are x = -1, x = 0, x = 3/8, and x = 1. These points are crucial for further analysis to determine the function's behavior within the given interval.

Analyzing Critical Points on the Interval [-1, 1]

Having identified the critical points of the function f(x) = 12x^4 - 6x^3 on the interval [-1, 1], which are x = -1, x = 0, x = 3/8, and x = 1, the next step is to analyze these points to determine whether they correspond to local maxima, local minima, or neither. This can be achieved through various methods, including the first derivative test and the second derivative test. Additionally, we must consider the endpoints of the interval to find the absolute maximum and minimum values of the function within the specified domain.

To begin, we can use the first derivative test. The first derivative test involves examining the sign of the first derivative, f'(x), in the intervals defined by the critical points. If f'(x) changes from negative to positive at a critical point, then the function has a local minimum at that point. Conversely, if f'(x) changes from positive to negative, the function has a local maximum. If f'(x) does not change sign, the point is neither a local maximum nor a local minimum.

Recall that the first derivative is f'(x) = 48x^3 - 18x^2 = 6x^2(8x - 3). We will analyze the sign of f'(x) in the intervals (-1, 0), (0, 3/8), and (3/8, 1).

  1. Interval (-1, 0): Choose a test point, such as x = -1/2. Then: f'(-1/2) = 6(-1/2)^2(8(-1/2) - 3) = 6(1/4)(-4 - 3) = (3/2)(-7) = -21/2 Since f'(-1/2) < 0, the function is decreasing in this interval.
  2. Interval (0, 3/8): Choose a test point, such as x = 1/4. Then: f'(1/4) = 6(1/4)^2(8(1/4) - 3) = 6(1/16)(2 - 3) = (3/8)(-1) = -3/8 Since f'(1/4) < 0, the function is also decreasing in this interval.
  3. Interval (3/8, 1): Choose a test point, such as x = 1/2. Then: f'(1/2) = 6(1/2)^2(8(1/2) - 3) = 6(1/4)(4 - 3) = (3/2)(1) = 3/2 Since f'(1/2) > 0, the function is increasing in this interval.

From the first derivative test, we can see that the function decreases on (-1, 0) and (0, 3/8), and increases on (3/8, 1). This indicates that there is a local minimum at x = 3/8. The derivative does not change sign at x = 0, so it is neither a local maximum nor a local minimum. To confirm these findings and to classify the endpoints, we can also use the second derivative test.

Applying the Second Derivative Test

The second derivative test is another method used to determine whether a critical point is a local maximum or local minimum. This test involves finding the second derivative of the function, f''(x), and evaluating it at the critical points. If f''(x) > 0 at a critical point, the function has a local minimum at that point. If f''(x) < 0, the function has a local maximum. If f''(x) = 0, the test is inconclusive, and we must use other methods, such as the first derivative test, to determine the nature of the critical point.

To apply the second derivative test to the function f(x) = 12x^4 - 6x^3, we first need to find the second derivative, f''(x). We already have the first derivative, f'(x) = 48x^3 - 18x^2. Now, we differentiate f'(x) to find f''(x):

f'(x) = 48x^3 - 18x^2

Applying the power rule again, we get:

f''(x) = 48(3x^2) - 18(2x)

Simplifying this expression, we obtain:

f''(x) = 144x^2 - 36x

Now that we have the second derivative, we evaluate it at the critical points we found earlier: x = 0 and x = 3/8.

  1. Evaluate f''(x) at x = 0: f''(0) = 144(0)^2 - 36(0) = 0 Since f''(0) = 0, the second derivative test is inconclusive at x = 0. We need to refer back to the first derivative test, which indicated that x = 0 is neither a local maximum nor a local minimum.
  2. Evaluate f''(x) at x = 3/8: f''(3/8) = 144(3/8)^2 - 36(3/8) = 144(9/64) - 108/8 Simplifying this expression: f''(3/8) = (144 * 9) / 64 - 108/8 = 1296/64 - 108/8 = 20.25 - 13.5 = 6.75 Since f''(3/8) = 6.75 > 0, the function has a local minimum at x = 3/8.

The second derivative test confirms that there is a local minimum at x = 3/8. To find the local minimum value, we substitute x = 3/8 into the original function f(x):

f(3/8) = 12(3/8)^4 - 6(3/8)^3

Calculating this value:

f(3/8) = 12(81/4096) - 6(27/512) = (12 * 81) / 4096 - (6 * 27) / 512 = 972/4096 - 162/512

To subtract these fractions, we need a common denominator. Since 4096 = 8 * 512, we multiply the second fraction by 8/8:

f(3/8) = 972/4096 - (162 * 8) / (512 * 8) = 972/4096 - 1296/4096 = -324/4096

Simplifying the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 108, we get:

f(3/8) = -324/4096 = -81/1024

Thus, the local minimum value of the function is -81/1024, which occurs at x = 3/8. This result, derived from the second derivative test, corroborates our earlier findings from the first derivative test.

Determining Global Extrema on [-1, 1]

To complete the analysis of the function f(x) = 12x^4 - 6x^3 on the interval [-1, 1], we need to determine the global (absolute) extrema. Global extrema are the highest and lowest values of the function over the entire interval. To find these, we evaluate the function at the critical points within the interval and at the endpoints of the interval. We have already identified the critical points as x = -1, x = 0, x = 3/8, and x = 1.

We have already calculated the local minimum value at x = 3/8, which is f(3/8) = -81/1024. Now, we need to evaluate the function at the endpoints x = -1 and x = 1, and at the other critical point x = 0.

  1. Evaluate f(x) at x = -1: f(-1) = 12(-1)^4 - 6(-1)^3 = 12(1) - 6(-1) = 12 + 6 = 18
  2. Evaluate f(x) at x = 0: f(0) = 12(0)^4 - 6(0)^3 = 0
  3. Evaluate f(x) at x = 1: f(1) = 12(1)^4 - 6(1)^3 = 12 - 6 = 6

Now, we compare the values of f(x) at these points:

  • f(-1) = 18
  • f(0) = 0
  • f(3/8) = -81/1024 ≈ -0.0791
  • f(1) = 6

From these values, we can determine the global extrema:

  • Global Maximum: The maximum value of the function on the interval [-1, 1] is 18, which occurs at x = -1.
  • Global Minimum: The minimum value of the function on the interval [-1, 1] is -81/1024, which occurs at x = 3/8.

Therefore, the global maximum of f(x) = 12x^4 - 6x^3 on the interval [-1, 1] is 18, and the global minimum is -81/1024. This analysis provides a complete understanding of the function's behavior within the given interval, including its critical points, local extrema, and global extrema.

Conclusion

In summary, for the function f(x) = 12x^4 - 6x^3 on the interval [-1, 1], we identified the critical points by finding where the first derivative is zero or undefined. We calculated the first derivative as f'(x) = 48x^3 - 18x^2 and found the critical points to be x = 0 and x = 3/8. Additionally, we considered the endpoints of the interval, x = -1 and x = 1.

Using the first derivative test, we analyzed the intervals determined by these critical points to understand the function's increasing and decreasing behavior. The analysis showed a local minimum at x = 3/8. To confirm this, we applied the second derivative test. The second derivative, f''(x) = 144x^2 - 36x, was evaluated at the critical points. The second derivative test confirmed the local minimum at x = 3/8 and was inconclusive at x = 0, which we determined to be neither a local maximum nor a local minimum based on the first derivative test.

The local minimum value was found to be f(3/8) = -81/1024. To determine the global extrema on the interval [-1, 1], we evaluated the function at the critical points and the endpoints. The results showed a global maximum of 18 at x = -1 and a global minimum of -81/1024 at x = 3/8.

This comprehensive analysis of the function f(x) = 12x^4 - 6x^3 on the interval [-1, 1] provides a clear understanding of its behavior, including the locations and values of its local and global extrema. This process illustrates the importance of using calculus techniques to analyze functions, which has broad applications in various fields such as physics, engineering, and economics. By mastering these techniques, one can gain valuable insights into the properties and behavior of functions, leading to informed decision-making and problem-solving in practical scenarios.