Extrema Of F(x) = X^5 - 4x^3 + X - 1 Analysis And Solution

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Deciphering the nature of extrema for a given function is a fundamental task in calculus, providing insights into the function's behavior and overall shape. In this comprehensive analysis, we embark on a journey to determine the extrema of the function f(x) = x^5 - 4x^3 + x - 1. We will meticulously investigate the existence of absolute maximum, absolute minimum, local maximum, and local minimum points, employing a combination of analytical techniques and graphical interpretations.

Delving into the Realm of Extrema: Absolute vs. Local

Before we embark on our exploration, it's crucial to establish a clear understanding of the different types of extrema. Extrema, in general, refer to the maximum and minimum values of a function, either within a specific interval (local extrema) or across the function's entire domain (absolute extrema).

  • Absolute Maximum: The absolute maximum represents the highest value attained by the function across its entire domain. It's the ultimate peak of the function's graph.
  • Absolute Minimum: Conversely, the absolute minimum signifies the lowest value attained by the function across its entire domain. It's the deepest valley in the function's graph.
  • Local Maximum: A local maximum represents a peak within a specific interval of the function's domain. It's a point where the function's value is higher than its immediate neighbors, but not necessarily the highest point overall.
  • Local Minimum: Analogous to the local maximum, a local minimum represents a valley within a specific interval of the function's domain. It's a point where the function's value is lower than its immediate neighbors, but not necessarily the lowest point overall.

Unveiling the Extrema: A Multifaceted Approach

To determine the extrema of f(x) = x^5 - 4x^3 + x - 1, we will employ a multifaceted approach, combining analytical techniques with graphical interpretations. This will allow us to gain a comprehensive understanding of the function's behavior and identify all its extrema.

1. The Power of Calculus: Derivatives to the Rescue

The cornerstone of our analytical approach lies in the power of calculus, specifically the concept of derivatives. The first derivative of a function, denoted as f'(x), provides crucial information about the function's slope and rate of change. Critical points, where f'(x) = 0 or f'(x) is undefined, are potential locations of local extrema.

The second derivative, f''(x), further refines our analysis. It reveals the concavity of the function, indicating whether the graph is curving upwards (concave up) or downwards (concave down). This information helps us distinguish between local maxima and local minima.

Let's embark on the process of finding the derivatives of our function:

  • f(x) = x^5 - 4x^3 + x - 1
  • f'(x) = 5x^4 - 12x^2 + 1
  • f''(x) = 20x^3 - 24x

2. Unmasking Critical Points: Where the Slope Vanishes

To identify potential local extrema, we need to find the critical points of the function. These are the points where the first derivative, f'(x), equals zero or is undefined. In our case, f'(x) = 5x^4 - 12x^2 + 1. This is a quartic equation, which can be solved by making a substitution y = x^2, transforming it into a quadratic equation: 5y^2 - 12y + 1 = 0.

Solving this quadratic equation using the quadratic formula, we get:

  • y = (12 ± √(12^2 - 4 * 5 * 1)) / (2 * 5)
  • y = (12 ± √124) / 10
  • y = (6 ± √31) / 5

Since y = x^2, we can find the values of x by taking the square root of both solutions for y:

  • x = ±√((6 + √31) / 5)
  • x = ±√((6 - √31) / 5)

Thus, we have four critical points:

  • x₁ = √((6 + √31) / 5) ˜ 1.531
  • x₂ = -√((6 + √31) / 5) ˜ -1.531
  • x₃ = √((6 - √31) / 5) ˜ 0.291
  • x₄ = -√((6 - √31) / 5) ˜ -0.291

3. The Second Derivative Test: Distinguishing Maxima and Minima

Now that we have identified the critical points, we can employ the second derivative test to determine whether they correspond to local maxima or local minima. The second derivative, f''(x), tells us about the concavity of the function at these points.

  • If f''(x) > 0 at a critical point, the function is concave up, indicating a local minimum.
  • If f''(x) < 0 at a critical point, the function is concave down, indicating a local maximum.
  • If f''(x) = 0 at a critical point, the test is inconclusive, and further analysis is required.

Let's evaluate the second derivative, f''(x) = 20x^3 - 24x, at each critical point:

  • f''(x₁) = f''(1.531) ˜ 20(1.531)^3 - 24(1.531) ˜ 31.3 > 0 (Local Minimum)
  • f''(x₂) = f''(-1.531) ˜ 20(-1.531)^3 - 24(-1.531) ˜ -31.3 < 0 (Local Maximum)
  • f''(x₃) = f''(0.291) ˜ 20(0.291)^3 - 24(0.291) ˜ -6.7 < 0 (Local Maximum)
  • f''(x₄) = f''(-0.291) ˜ 20(-0.291)^3 - 24(-0.291) ˜ 6.7 > 0 (Local Minimum)

Therefore, we have identified two local minima at x₁ ˜ 1.531 and x₄ ˜ -0.291, and two local maxima at x₂ ˜ -1.531 and x₃ ˜ 0.291.

4. The Grand View: Graphical Interpretation

To gain a holistic understanding of the function's behavior, let's turn to graphical interpretation. Plotting the graph of f(x) = x^5 - 4x^3 + x - 1 reveals the overall shape and the location of its extrema.

[Insert Graph of f(x) = x^5 - 4x^3 + x - 1]

The graph visually confirms our analytical findings. We can clearly observe the two local minima and two local maxima, as well as the unbounded nature of the function, extending infinitely upwards and downwards.

5. The Quest for Absolute Extrema: Unbounded Behavior

Now, let's address the question of absolute extrema. Absolute extrema, as we discussed earlier, represent the highest and lowest values attained by the function across its entire domain. However, for our function, f(x) = x^5 - 4x^3 + x - 1, the story is different.

As x approaches positive infinity (x → ∞), the function also tends towards positive infinity (f(x) → ∞). This indicates that there is no absolute maximum.

Conversely, as x approaches negative infinity (x → -∞), the function tends towards negative infinity (f(x) → -∞). This implies that there is no absolute minimum.

This unbounded behavior is characteristic of polynomial functions with odd degrees, such as our quintic function (x^5).

The Verdict: Extrema Unveiled

In conclusion, our comprehensive analysis of f(x) = x^5 - 4x^3 + x - 1 reveals the following extrema:

  • Absolute Maximum: None
  • Absolute Minimum: None
  • Local Maximum: Two (at approximately x = -1.531 and x = 0.291)
  • Local Minimum: Two (at approximately x = 1.531 and x = -0.291)

Therefore, the correct answer is D. III and IV only.

This exploration highlights the power of calculus and graphical interpretation in unraveling the intricacies of function behavior and identifying extrema. By combining analytical techniques with visual representations, we gain a profound understanding of the function's characteristics and its overall shape.