How to Find Relative Minimum and Maximum Without A Calculator

What Are Relative Minimum and Maximum?

In calculus, a relative (or local) minimum is a point on a function where the function values are lower than all nearby points. Similarly, a relative maximum is a point where the function values are higher than all nearby points. These points are important in optimization problems and understanding the behavior of functions.

To find these points without a calculator, we use calculus techniques that analyze the function's derivatives. The two main methods are:

1. First Derivative Test
2. Second Derivative Test

First Derivative Test

The first derivative test is a straightforward method to identify relative minima and maxima. 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. Use the following rules:
   • If f'(x) changes from negative to positive, it's a relative minimum.
   • If f'(x) changes from positive to negative, it's a relative maximum.
   • If f'(x) doesn't change sign, it's neither.

Example

Consider the function f(x) = x³ - 3x². Let's find its critical points and classify them using the first derivative test.

1. Find f'(x): 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 (e.g., x = -1): f'(-1) = 3(-1)² - 6(-1) = 3 + 6 = 9 > 0
   • For 0 < x < 2 (e.g., x = 1): f'(1) = 3(1)² - 6(1) = 3 - 6 = -3 < 0
   • For x > 2 (e.g., x = 3): f'(3) = 3(3)² - 6(3) = 27 - 18 = 9 > 0
4. Conclusion:
   • At x = 0: f'(x) changes from + to - → relative maximum
   • At x = 2: f'(x) changes from - to + → relative minimum

Second Derivative Test

The second derivative test provides a quicker way to classify critical points when the first derivative test is too cumbersome. Here's how it works:

1. Find all critical points by solving f'(x) = 0 or where f'(x) is undefined.
2. Compute the second derivative f''(x).
3. Evaluate f''(x) at each critical point:
   • If f''(x) > 0, it's a relative minimum.
   • If f''(x) < 0, it's a relative maximum.
   • If f''(x) = 0, the test is inconclusive.

Example

Using the same function f(x) = x³ - 3x², let's apply the second derivative test.

1. Find f''(x): f''(x) = 6x - 6
2. Evaluate at critical points:
   • At x = 0: f''(0) = 6(0) - 6 = -6 < 0 → relative maximum
   • At x = 2: f''(2) = 6(2) - 6 = 6 > 0 → relative minimum

Comparison of Methods

Both methods have their advantages and disadvantages:

The first derivative test is more reliable but requires more work. The second derivative test is quicker but has limitations. In practice, you might use both methods to confirm results.

Worked Example

Let's find and classify all critical points for the function f(x) = x⁴ - 4x³.

1. Find f'(x): f'(x) = 4x³ - 12x²
2. Set f'(x) = 0: 4x³ - 12x² = 0 → 4x²(x - 3) = 0 → x = 0 or x = 3
3. First Derivative Test:
   • For x < 0 (e.g., x = -1): f'(-1) = 4(-1)³ - 12(-1)² = -4 - 12 = -16 < 0
   • For 0 < x < 3 (e.g., x = 1): f'(1) = 4(1)³ - 12(1)² = 4 - 12 = -8 < 0
   • For x > 3 (e.g., x = 4): f'(4) = 4(4)³ - 12(4)² = 256 - 192 = 64 > 0

   At x = 0: f'(x) changes from - to - → neither

   At x = 3: f'(x) changes from - to + → relative minimum

4. Second Derivative Test:
   • Find f''(x): f''(x) = 12x² - 24x
   • At x = 0: f''(0) = 0 → inconclusive
   • At x = 3: f''(3) = 12(9) - 24(3) = 108 - 72 = 36 > 0 → relative minimum

Conclusion: The function has a relative minimum at x = 3. The point x = 0 is neither a minimum nor maximum.