Problem 29 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=3 x^{4}-4 x^{3}-\frac{9}{2} x^{2}+\frac{1}{2} $$

Step-by-Step Solution

Answer

Relative maximum at \( x = 0 \) and relative minimum at \( x = 1.5 \).

Step 1: Find the First Derivative

To use the Second Derivative Test, we first need to find the critical points, which require the first derivative of the function. Differentiate the function, \( f(x) = 3x^4 - 4x^3 - \frac{9}{2}x^2 + \frac{1}{2} \), to find the first derivative \( f'(x) \).\[ f'(x) = 12x^3 - 12x^2 - 9x \]

Step 2: Find the Critical Points

Set the first derivative equal to zero to find the critical points.\[ 12x^3 - 12x^2 - 9x = 0 \]Factor the equation:\[ 3x(4x^2 - 4x - 3) = 0 \]This results in:\[ 3x = 0 \quad \text{or} \quad 4x^2 - 4x - 3 = 0 \]Solving for \( x \), the first equation gives \( x = 0 \). The quadratic equation \( 4x^2 - 4x - 3 = 0 \) is solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 4, b = -4, \) and \( c = -3 \). After solving,\[ x = 1.5 \quad \text{or} \quad x = -0.5 \]

Step 3: Find the Second Derivative

Differentiate the first derivative \( f'(x) \) to find the second derivative \( f''(x) \), which is needed for the second derivative test.\[ f''(x) = 36x^2 - 24x - 9 \]

Step 4: Apply the Second Derivative Test

Evaluate the second derivative at each critical point to determine the nature of each critical point:1. **At \( x = 0 \)**: \[ f''(0) = 36(0)^2 - 24(0) - 9 = -9 \] Since \( f''(0) < 0 \), \( f(x) \) has a relative maximum at \( x = 0 \).2. **At \( x = 1.5 \)**: \[ f''(1.5) = 36(1.5)^2 - 24(1.5) - 9 = 27 \] Since \( f''(1.5) > 0 \), \( f(x) \) has a relative minimum at \( x = 1.5 \).3. **At \( x = -0.5 \)**: \[ f''(-0.5) = 36(-0.5)^2 - 24(-0.5) - 9 = 0 \] Since \( f''(-0.5) = 0 \), the test is inconclusive for \( x = -0.5 \).

Key Concepts

Critical Points

Critical points are essential in understanding where a function might reach a peak or valley. They are key indicators for analyzing potential maximum or minimum values. Take a function, differentiate it, and you identify these points by setting the first derivative to zero. These zero points of the derivative function tell us where the slope of the original function is horizontal, meaning a relative extremum might exist there.
In our exercise, the first derivative was set to zero as follows:

• The derivative,\( f'(x) = 12x^3 - 12x^2 - 9x \), is set equal to zero.
• This allows us to solve for the critical points: \( x = 0, 1.5, \) and \( -0.5 \).

By working through these calculations, we identify the points where the function might possess relative maximum or minimum values.

First Derivative

The first derivative of a function is crucial in determining the behavior of that function. It represents the rate of change or slope of the original function. In the context of relative extrema, finding where the first derivative equals zero is the first step in the Second Derivative Test. Why zero? Because at critical points, where the derivative is zero, the function's slope is flat, possibly indicating a peak or trough.
Taking the original function, \( f(x) = 3x^4 - 4x^3 - \frac{9}{2}x^2 + \frac{1}{2} \), and differentiating, we find:

• \( f'(x) = 12x^3 - 12x^2 - 9x \)

This first derivative is strategically set to zero in order to find potential turning points on the function's graph.

Second Derivative

The second derivative provides insights into the concavity of a function, which helps determine the nature of the critical points found using the first derivative. If the second derivative is positive at a critical point, the function is concave up, indicating a relative minimum. If negative, the function is concave down, suggesting a relative maximum.
For our function, the second derivative is:

• \( f''(x) = 36x^2 - 24x - 9 \)

By evaluating this at each critical point, we can decide the nature of these points:

• At \( x = 0 \), \( f''(0) = -9 \), suggesting a relative maximum.
• At \( x = 1.5 \), \( f''(1.5) = 27 \), indicating a relative minimum.
• At \( x = -0.5 \), \( f''(-0.5) = 0 \), which is inconclusive, so further analysis would be needed for this point.

Relative Extrema

Relative extrema are the peaks and valleys on a graph of a function, within a certain interval. These are the largest (maximum) or smallest (minimum) values compared to points around them. The Second Derivative Test helps us determine whether a critical point is a relative maximum or minimum by taking the second derivative and evaluating it at the critical points.
In our problem, using the second derivative at critical points:

• \( x = 0 \): Relative Maximum because \( f''(0) < 0 \).
• \( x = 1.5 \): Relative Minimum since \( f''(1.5) > 0 \).
• \( x = -0.5 \): This is inconclusive as \( f''(-0.5) = 0 \). Additional tests would be necessary to determine its nature.

This understanding allows the identification of local maximas and minimas, vital in both theoretical and applied calculus settings.