Relative Extrema - Algebra-2

1. Fundamental Concepts

Definition: Relative extrema (maxima and minima) are the highest or lowest points in a local region of a polynomial function.
Critical Points: Points where the derivative of the polynomial is zero or undefined.
First Derivative Test: Determines whether a critical point is a relative maximum, minimum, or neither by examining the sign changes of the first derivative around the critical point.

2. Key Concepts

First Derivative Test: If $$ f'(x) $$ changes from positive to negative at $$ c $$, then $$ f(c) $$ is a relative maximum. If $$ f'(x) $$ changes from negative to positive at $$ c $$, then $$ f(c) $$ is a relative minimum.

Second Derivative Test: If $$ f''(c) > 0 $$, then $$ f(c) $$ is a relative minimum. If $$ f''(c) < 0 $$, then $$ f(c) $$ is a relative maximum. If $$ f''(c) = 0 $$, the test is inconclusive.

Application: Relative extrema are used in optimization problems, such as finding the maximum volume of a box or the minimum cost of production.

3. Examples

Example 1 (Basic)

Problem: Find the relative extrema of the function $$ f(x) = x^3 - 3x^2 + 4 $$.

Step-by-Step Solution:

Find the first derivative: $$ f'(x) = 3x^2 - 6x $$.
Set the first derivative equal to zero to find critical points: $$ 3x^2 - 6x = 0 $$. Factor: $$ 3x(x - 2) = 0 $$. So, $$ x = 0 $$ and $$ x = 2 $$.
Use the first derivative test:
- For $$ x < 0 $$, $$ f'(x) > 0 $$ (positive).
- For $$ 0 < x < 2 $$, $$ f'(x) < 0 $$ (negative).
- For $$ x > 2 $$, $$ f'(x) > 0 $$ (positive).
Therefore, $$ f(0) = 4 $$ is a relative maximum, and $$ f(2) = 0 $$ is a relative minimum.

Validation: Substitute $$ x = 0 $$ and $$ x = 2 $$ into the original function to verify the values.

Example 2 (Intermediate)

Problem: Find the relative extrema of the function $$ g(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 $$.

Step-by-Step Solution:

Find the first derivative: $$ g'(x) = 4x^3 - 12x^2 + 12x - 4 $$.
Set the first derivative equal to zero to find critical points: $$ 4x^3 - 12x^2 + 12x - 4 = 0 $$. Factor: $$ 4(x-1)^3 = 0 $$. So, $$ x = 1 $$.
Use the second derivative test:
- Find the second derivative: $$ g''(x) = 12x^2 - 24x + 12 $$.
- Evaluate the second derivative at $$ x = 1 $$: $$ g''(1) = 12(1)^2 - 24(1) + 12 = 0 $$. The second derivative test is inconclusive.
- Use the first derivative test:
  - For $$ x < 1 $$, $$ g'(x) < 0 $$ (negative).
  - For $$ x > 1 $$, $$ g'(x) > 0 $$ (positive).
  Therefore, $$ g(1) = 0 $$ is a relative minimum.

Validation: Substitute $$ x = 1 $$ into the original function to verify the value.

4. Problem-Solving Techniques

Graphical Analysis: Use graphing calculators or software to visualize the function and identify potential extrema.
Sign Analysis: Create a sign chart for the first derivative to determine where the function is increasing or decreasing.
Second Derivative Test: Use the second derivative to confirm the nature of the critical points when the first derivative test is inconclusive.