How to Find Relative Max and Min Without Calculator

What Are Relative Extrema?

Relative extrema are points on a function where the function values are either higher or lower than their immediate neighbors. These are also called local maxima and minima.

Key characteristics of relative extrema:

• Relative maxima: A point where the function value is greater than all nearby points
• Relative minima: A point where the function value is less than all nearby points
• Critical points: Points where the derivative is zero or undefined
• Not all critical points are extrema - some are points of inflection

First Derivative Test

The first derivative test is the most common method for identifying relative extrema. Here's how it works:

1. Find all critical points by solving f'(x) = 0 or where f'(x) is undefined
2. Determine the sign of f'(x) on either side of each critical point
3. Apply the following rules:
   • If f'(x) changes from positive to negative, it's a relative maximum
   • If f'(x) changes from negative to positive, it's a relative minimum
   • If f'(x) doesn't change sign, it's not an extremum

Example Using First Derivative Test

Consider the function f(x) = x³ - 3x² + 4

1. Find the derivative: f'(x) = 3x² - 6x
2. Set f'(x) = 0: 3x² - 6x = 0 → 3x(x - 2) = 0 → x = 0 or x = 2
3. Test intervals:
   • For x < 0: f'(x) = positive (e.g., x = -1 → f'(-1) = 9 > 0)
   • For 0 < x < 2: f'(x) = negative (e.g., x = 1 → f'(1) = -3 < 0)
   • For x > 2: f'(x) = positive (e.g., x = 3 → f'(3) = 9 > 0)
4. Conclusion:
   • At x = 0: positive to negative → relative maximum
   • At x = 2: negative to positive → relative minimum

Second Derivative Test

The second derivative test provides a quicker method when the first derivative test is too cumbersome:

1. Find all critical points as before
2. Find the second derivative f''(x)
3. Evaluate f''(x) at each critical point:
   • If f''(x) > 0 → relative minimum
   • If f''(x) < 0 → relative maximum
   • If f''(x) = 0 → test is inconclusive

Example Using Second Derivative Test

Using the same function f(x) = x³ - 3x² + 4

1. First derivative: f'(x) = 3x² - 6x
2. Critical points: x = 0 and x = 2
3. Second derivative: f''(x) = 6x - 6
4. Evaluate at critical points:
   • At x = 0: f''(0) = -6 < 0 → relative maximum
   • At x = 2: f''(2) = 6 > 0 → relative minimum

Comparison of Methods

The first derivative test is generally preferred because it's more reliable, though the second derivative test can be quicker when it works.

Worked Example

Find all relative extrema for f(x) = x⁴ - 4x³ + 4

Solution

1. Find the first derivative: f'(x) = 4x³ - 12x²
2. Find critical points: 4x³ - 12x² = 0 → 4x²(x - 3) = 0 → x = 0 or x = 3
3. Find the second derivative: f''(x) = 12x² - 24x
4. Evaluate at critical points:
   • At x = 0: f''(0) = 0 → inconclusive
   • At x = 3: f''(3) = 108 - 72 = 36 > 0 → relative minimum
5. For x = 0, use first derivative test:
   • For x < 0: f'(x) = positive (e.g., x = -1 → f'(-1) = -16 < 0)
   • For 0 < x < 3: f'(x) = negative (e.g., x = 1 → f'(1) = -8 < 0)
   • For x > 3: f'(x) = positive (e.g., x = 4 → f'(4) = 16 > 0)

   Since f'(x) doesn't change sign around x = 0, it's not an extremum (point of inflection)

6. Conclusion: Only relative minimum at x = 3