Critical Points And Local Minima Of F(x) = 12x^4 - 6x^3

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Introduction: Exploring Critical Points and Local Minima

In the realm of calculus, understanding the behavior of functions is paramount. Critical points, those special locations where a function's derivative vanishes or is undefined, hold the key to unlocking a function's local extrema – its peaks and valleys. The second derivative test provides a powerful tool for discerning the nature of these critical points, helping us identify local minima, where the function attains its lowest values within a specific neighborhood. In this article, we embark on a journey to unravel the critical points and local minimum values of the function f(x) = 12x^4 - 6x^3, defined over the interval [-1, 1].

This exploration will not only solidify our grasp of fundamental calculus concepts but also showcase the practical application of these concepts in analyzing the behavior of functions. By meticulously examining the function's first and second derivatives, we will pinpoint the critical points and employ the second derivative test to definitively determine the local minimum values. This analytical process exemplifies the power of calculus in providing a comprehensive understanding of mathematical functions.

Finding the Critical Points: A Step-by-Step Approach

To embark on our quest to find the critical points of the function f(x) = 12x^4 - 6x^3, we must first delve into the realm of differential calculus. Critical points, as we've mentioned, are the points where the function's derivative either equals zero or is undefined. These points mark potential locations of local maxima, local minima, or saddle points, where the function's behavior undergoes a significant change. Our mission is to identify these critical junctures within the function's domain, the interval [-1, 1].

  1. Compute the First Derivative: The cornerstone of our approach lies in finding the first derivative of the function, denoted as f'(x). The derivative, in essence, quantifies the instantaneous rate of change of the function at any given point. Using the power rule of differentiation, we can systematically compute the derivative of f(x) = 12x^4 - 6x^3:

f'(x) = d/dx (12x^4 - 6x^3) = 48x^3 - 18x^2

The resulting expression, f'(x) = 48x^3 - 18x^2, represents the slope of the tangent line to the function's graph at any point x. This derivative serves as our guide in locating the critical points, where the tangent line becomes horizontal (slope equals zero) or the derivative is undefined.

  1. Set the First Derivative to Zero: To pinpoint the critical points, we set the first derivative, f'(x), equal to zero. This step stems from the fact that at local maxima and minima, the tangent line to the function's graph is horizontal, implying a zero slope. Thus, we solve the equation:

48x^3 - 18x^2 = 0

This equation is a cubic polynomial equation, which may seem daunting at first glance. However, we can employ algebraic techniques to simplify and solve it. Factoring out the common factor of 6x^2, we get:

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

This factored form reveals the roots of the equation, the values of x that make the expression equal to zero. The roots are x = 0 and x = 3/8.

  1. Identify Critical Points: The solutions we obtained, x = 0 and x = 3/8, represent the critical points of the function. These are the points within the interval [-1, 1] where the function's derivative is zero, indicating potential locations of local extrema. However, critical points can also arise where the derivative is undefined. In the case of our polynomial function, the derivative is defined for all real numbers, so we don't need to consider this scenario.

  2. Check Endpoints of the Interval: Since we are examining the function over the closed interval [-1, 1], we must also consider the endpoints of the interval as potential critical points. The endpoints, x = -1 and x = 1, represent the boundaries of our domain, and the function's behavior at these points is crucial in determining its overall extrema. Therefore, we include x = -1 and x = 1 in our list of critical points.

  3. List of Critical Points: In conclusion, the critical points of the function f(x) = 12x^4 - 6x^3 within the interval [-1, 1] are:

  • x = -1
  • x = 0
  • x = 3/8
  • x = 1

These critical points mark the locations where the function's behavior may transition from increasing to decreasing or vice versa. To definitively determine the nature of these critical points, whether they represent local minima, local maxima, or saddle points, we turn to the second derivative test.

Applying the Second Derivative Test: Unveiling Local Minima

The second derivative test serves as our compass in navigating the landscape of critical points. This powerful tool harnesses the information encoded in the function's second derivative to discern the nature of each critical point – whether it corresponds to a local minimum, a local maximum, or neither. Our focus here is to identify the local minimum values of the function f(x) = 12x^4 - 6x^3 within the interval [-1, 1].

  1. Compute the Second Derivative: The cornerstone of the second derivative test lies in computing the second derivative of the function, denoted as f''(x). The second derivative, in essence, quantifies the concavity of the function – whether the graph curves upwards (concave up) or downwards (concave down). To find the second derivative, we differentiate the first derivative, f'(x) = 48x^3 - 18x^2:

f''(x) = d/dx (48x^3 - 18x^2) = 144x^2 - 36x

The resulting expression, f''(x) = 144x^2 - 36x, represents the rate of change of the slope of the tangent line to the function's graph. This information is crucial in determining the nature of the critical points.

  1. Evaluate the Second Derivative at Critical Points: To apply the second derivative test, we evaluate the second derivative, f''(x), at each of the critical points we identified earlier: x = -1, x = 0, x = 3/8, and x = 1. The sign of the second derivative at a critical point provides valuable insight into the function's behavior:
  • If f''(x) > 0, the function is concave up at that point, indicating a local minimum.
  • If f''(x) < 0, the function is concave down at that point, indicating a local maximum.
  • If f''(x) = 0, the test is inconclusive, and further analysis is needed.

Let's evaluate the second derivative at each critical point:

  • f''(-1) = 144(-1)^2 - 36(-1) = 144 + 36 = 180 > 0
  • f''(0) = 144(0)^2 - 36(0) = 0
  • f''(3/8) = 144(3/8)^2 - 36(3/8) = 144(9/64) - 108/8 = 81/4 - 108/8 = 162/8 - 108/8 = 54/8 > 0
  • f''(1) = 144(1)^2 - 36(1) = 144 - 36 = 108 > 0
  1. Identify Local Minima: Based on the second derivative test, we can now identify the local minima of the function. At the critical points where f''(x) > 0, the function is concave up, indicating a local minimum. From our evaluations, we see that:
  • f''(-1) = 180 > 0, suggesting a local minimum at x = -1
  • f''(3/8) = 54/8 > 0, suggesting a local minimum at x = 3/8
  • f''(1) = 108 > 0, suggesting a local minimum at x = 1

The second derivative test is inconclusive at x = 0 since f''(0) = 0. This means we need to employ other methods, such as examining the sign of the first derivative around x = 0, to determine the function's behavior at this point.

  1. Calculate Local Minimum Values: To find the actual local minimum values, we substitute the x-values of the local minima into the original function, f(x) = 12x^4 - 6x^3:
  • f(-1) = 12(-1)^4 - 6(-1)^3 = 12 + 6 = 18
  • f(3/8) = 12(3/8)^4 - 6(3/8)^3 = 12(81/4096) - 6(27/512) = 972/4096 - 162/512 = 972/4096 - 1296/4096 = -324/4096 = -81/1024
  • f(1) = 12(1)^4 - 6(1)^3 = 12 - 6 = 6

Therefore, the local minimum values of the function f(x) = 12x^4 - 6x^3 within the interval [-1, 1] are:

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

Conclusion: Critical Points and Local Minima Unveiled

In this comprehensive exploration, we embarked on a journey to unveil the critical points and local minimum values of the function f(x) = 12x^4 - 6x^3, defined over the interval [-1, 1]. We meticulously computed the first and second derivatives, systematically identified the critical points, and masterfully employed the second derivative test to discern the nature of these critical points.

Our analysis revealed that the critical points of the function within the interval [-1, 1] are x = -1, x = 0, x = 3/8, and x = 1. By applying the second derivative test, we definitively identified local minima at x = -1, x = 3/8, and x = 1. Furthermore, we calculated the corresponding local minimum values, which are f(-1) = 18, f(3/8) = -81/1024, and f(1) = 6.

This analytical journey underscores the power of calculus in providing a profound understanding of the behavior of functions. By delving into the derivatives, we unlocked the secrets of critical points and local extrema, gaining valuable insights into the function's landscape. This knowledge equips us with the tools to effectively analyze and interpret mathematical models in various fields of science, engineering, and economics.

Final Answer

The critical points for the function f(x) = 12x^4 - 6x^3, x ∈ [-1, 1] are x = 0 and x = 3/8. The local minimum values were found using the second derivative test.