Problem 24 - Calculus A New Horizon

Question

Use any method to find the relative extrema of the function \(f\). $$f(x)=x^{4}+2 x^{3}$$

Step-by-Step Solution

Answer

Local maximum at \( x = -\frac{3}{2} \), no local extremum at \( x = 0 \).

Step 1: Find the First Derivative

To find the relative extrema, we first need to compute the first derivative of the function. The function we have is \( f(x) = x^4 + 2x^3 \). We apply the power rule for differentiation: the derivative of \( x^n \) is \( nx^{n-1} \). Therefore, the first derivative is \( f'(x) = 4x^3 + 6x^2 \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 0 \): \( 4x^3 + 6x^2 = 0 \). Factor out common terms: \( 2x^2(2x + 3) = 0 \). The critical points are then \( x = 0 \) and \( x = -\frac{3}{2} \).

Step 3: Use the Second Derivative Test

To determine the nature of each critical point, we use the second derivative test. First, compute the second derivative: \( f''(x) = 12x^2 + 12x = 12x(x + 1) \). Evaluate the second derivative at the critical points. For \( x = 0 \), \( f''(0) = 0 \), which is inconclusive with the second derivative test. For \( x = -\frac{3}{2} \), \( f''(-\frac{3}{2}) = 12(-\frac{3}{2})((-\frac{3}{2})+1) \). Calculate to see if \( f''(-\frac{3}{2}) \) is greater or less than zero.

Step 4: Determine Types of Extrema

For \( x = -\frac{3}{2} \), computing gives a negative value for \( f''(-\frac{3}{2}) \), indicating a local maximum since the second derivative is less than zero. Since \( f''(0) = 0 \) is inconclusive, further analysis is needed (for example, using a sign-change analysis for \( f'(x) \) around \( x = 0 \) indicates there is no local extremum since \( f'(x) \) changes sign from negative to positive).

Key Concepts

Relative Extrema

In calculus, relative extrema refer to the points where a function reaches a minimum or maximum value within a specific interval. These aren't the absolute highest or lowest points across the entire domain but rather peaks and valleys in localized areas. Identifying these points is crucial because they help us understand the behavior of a function. Relative extrema occur at critical points, which are found by examining the function's derivatives. By pinpointing these spots, we learn where a function's graph shifts direction, either turning up into a peak or down into a valley. This concept helps in studying the function's overall shape and improving predictions about its behavior in different contexts.

Critical Points

Critical points are specific values of the variable, often denoted as \( x \), where the derivative of a function is either zero or undefined. These points play a pivotal role in calculus because they are potential candidates for identifying relative extrema. To find them, you first take the function's first derivative and solve for the values that make this derivative zero.

• Zero: The slope of the tangent is flat, often indicating potential peaks or valleys.
• Undefined: Indicates possible corners or cusp points.

For the given function, the first derivative was set to zero, \( f'(x) = 4x^3 + 6x^2 = 0 \), which churned out the critical points \( x = 0 \) and \( x = -\frac{3}{2} \). These critical points help us explore the possibility of relative extrema.

Second Derivative Test

The second derivative test is utilized to determine the nature of critical points identified by the first derivative test. By finding the second derivative \( f''(x) \) of the original function, you can evaluate the concavity at each critical point. This information helps decide whether each point is a local minimum, maximum, or if the test is inconclusive. When applying the test, observe the following:

• If \( f''(x) > 0 \), the function is concave upward at the critical point, suggesting a local minimum.
• If \( f''(x) < 0 \), the function is concave downward at the critical point, indicating a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, necessitating additional methods to determine the nature of the critical point.

For this exercise, at \( x = -\frac{3}{2} \), \( f''(x) \) gave a negative value, hence a local maximum was identified. For \( x = 0 \), \( f''(x) = 0 \), leading to inconclusive results requiring further analysis.

Factorization

Factorization is a technique used in calculus and algebra to break down complex expressions into simpler components, making it easier to solve equations, particularly when finding roots or simplifying derivatives. By expressing a polynomial as a product of its factors, solutions to equations can be revealed or simplified. This strategy is crucial when working with derivatives to find critical points.

• Extract common factors from all terms to simplify expressions.
• Use identities or methods like grouping for more complex polynomials.
• Once in factored form, set each factor equal to zero to find solutions.

In the given solution, after finding the derivative \( f'(x) = 4x^3 + 6x^2 \), factorization was applied, written as \( 2x^2(2x + 3) = 0 \), revealing the critical points \( x = 0 \) and \( x = -\frac{3}{2} \). This step simplified the process of finding where the derivative equals zero, facilitating the analysis of potential relative extrema.