Derivatives And Extrema Analysis Of F(x) = 3x⁴ + 4x³ - 12x² + 2

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Introduction

In this article, we delve into the intricacies of the function f(x) = 3x⁴ + 4x³ - 12x² + 2. Our primary objective is to compute the first derivative, denoted as f'(x), and the second derivative, f''(x), and subsequently evaluate these derivatives at x = -2. Furthermore, we aim to analyze the behavior of the function at this critical point to determine whether it exhibits a relative maximum, a relative minimum, or neither. Understanding the nature of a function's derivatives and their implications for its extrema is a cornerstone of calculus and finds widespread applications in various fields, including optimization problems in engineering, economics, and computer science. This exploration will not only enhance our understanding of this specific function but also reinforce fundamental concepts in calculus.

Calculating the First Derivative f'(x)

To begin our analysis, we need to find the first derivative of the function f(x) = 3x⁴ + 4x³ - 12x² + 2. The first derivative, f'(x), provides valuable information about the slope of the tangent line to the curve at any given point. It helps us identify critical points where the function may have local maxima or minima. We will apply the power rule of differentiation, which states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. This rule is fundamental in calculus and allows us to efficiently differentiate polynomial functions.

Applying the power rule to each term of the function, we get:

  • The derivative of 3x⁴ is 12x³.
  • The derivative of 4x³ is 12x².
  • The derivative of -12x² is -24x.
  • The derivative of the constant 2 is 0.

Combining these results, we find the first derivative:

f'(x) = 12x³ + 12x² - 24x

Now that we have the first derivative, we can evaluate it at x = -2 to find the slope of the tangent line at that specific point. This value will be crucial in determining the function's behavior around x = -2.

Evaluating f'(-2)

Now that we have calculated the first derivative, f'(x) = 12x³ + 12x² - 24x, our next step is to evaluate it at x = -2. This will give us the slope of the tangent line to the curve of f(x) at the point where x = -2. This information is essential for determining whether the function is increasing or decreasing at that point, which is a key step in identifying local maxima and minima. By substituting x = -2 into the expression for f'(x), we can directly compute the value of the derivative at this specific point.

Substituting x = -2 into f'(x), we get:

f'(-2) = 12(-2)³ + 12(-2)² - 24(-2)

Let's break down the calculation:

  • (-2)³ = -8, so 12(-2)³ = 12(-8) = -96
  • (-2)² = 4, so 12(-2)² = 12(4) = 48
  • -24(-2) = 48

Now, we combine these results:

f'(-2) = -96 + 48 + 48 = 0

Therefore, f'(-2) = 0. This result is significant because it indicates that at x = -2, the tangent line to the curve is horizontal. This is a characteristic of critical points, which may be local maxima, local minima, or saddle points. To determine the exact nature of the critical point at x = -2, we need to further analyze the function, specifically by examining the second derivative.

Calculating the Second Derivative f''(x)

To gain deeper insights into the behavior of the function f(x) = 3x⁴ + 4x³ - 12x² + 2, we need to calculate the second derivative, denoted as f''(x). The second derivative provides information about the concavity of the function's curve. A positive second derivative indicates that the function is concave up (shaped like a U), while a negative second derivative indicates that the function is concave down (shaped like an upside-down U). The second derivative test is a powerful tool for determining whether a critical point is a local maximum, a local minimum, or neither. To find f''(x), we differentiate the first derivative, f'(x) = 12x³ + 12x² - 24x, with respect to x.

Again, we apply the power rule of differentiation to each term of f'(x):

  • The derivative of 12x³ is 36x².
  • The derivative of 12x² is 24x.
  • The derivative of -24x is -24.

Combining these results, we obtain the second derivative:

f''(x) = 36x² + 24x - 24

With the second derivative in hand, we can evaluate it at x = -2 to determine the concavity of the function at that point. This, combined with the information from the first derivative, will allow us to classify the critical point at x = -2.

Evaluating f''(-2)

Having computed the second derivative, f''(x) = 36x² + 24x - 24, we now evaluate it at x = -2. This will tell us about the concavity of the function f(x) at that specific point. The concavity, in conjunction with the information from the first derivative, is crucial for determining whether x = -2 corresponds to a local maximum, a local minimum, or neither. A positive value of f''(-2) indicates that the function is concave up at x = -2, suggesting a local minimum. Conversely, a negative value suggests a local maximum. If f''(-2) = 0, the second derivative test is inconclusive, and further analysis would be required.

Substituting x = -2 into the expression for f''(x), we get:

f''(-2) = 36(-2)² + 24(-2) - 24

Let's break down the calculation:

  • (-2)² = 4, so 36(-2)² = 36(4) = 144
  • 24(-2) = -48

Now, we combine these results:

f''(-2) = 144 - 48 - 24 = 72

Therefore, f''(-2) = 72. Since f''(-2) is positive, this indicates that the function is concave up at x = -2. Combined with our earlier finding that f'(-2) = 0, we can now confidently classify the critical point at x = -2.

Determining the Nature of the Relative Extrema

Now that we have computed both the first and second derivatives of the function f(x) = 3x⁴ + 4x³ - 12x² + 2 and evaluated them at x = -2, we can determine the nature of the relative extrema at this point. We found that f'(-2) = 0, which means that x = -2 is a critical point. Furthermore, we found that f''(-2) = 72, which is a positive value. The second derivative test states that if f'(c) = 0 and f''(c) > 0, then f(x) has a local minimum at x = c. In our case, c = -2, and the conditions of the second derivative test are met.

Therefore, we can conclude that the function f(x) = 3x⁴ + 4x³ - 12x² + 2 has a relative minimum at x = -2. This means that in the immediate vicinity of x = -2, the function reaches a minimum value compared to other points nearby. This is a crucial piece of information for understanding the overall behavior and shape of the function's graph.

Conclusion

In summary, we have thoroughly analyzed the function f(x) = 3x⁴ + 4x³ - 12x² + 2 by calculating its first and second derivatives and evaluating them at x = -2. We found that f'(-2) = 0, indicating a critical point, and f''(-2) = 72, a positive value, indicating that the function is concave up at x = -2. Applying the second derivative test, we confidently concluded that f(x) has a relative minimum at x = -2. This comprehensive analysis demonstrates the power of calculus in understanding the behavior of functions, particularly in identifying and classifying local extrema. This knowledge is invaluable in a wide range of applications, from optimization problems in various scientific and engineering disciplines to curve sketching and mathematical modeling.