Maxima and Minima

In this lesson you’ll learn how to find where a function reaches its highest and lowest values, both locally and on a closed interval. This is the foundation for optimization problems.

A critical point is where f’(x) = 0 or f’(x) doesn’t exist. These are the only places where a local max or min can happen. If the derivative exists and isn’t zero, the function is either going up or going down, so there’s no peak or valley.

Local Maximum f' + then f' −

Local Minimum f' − then f' +

At a local maximum (left), the function rises (f’ positive, green arrow) then falls (f’ negative, red arrow). The tangent line is horizontal at the peak. At a local minimum (right), the function falls then rises. The tangent line is horizontal at the valley.

Check the sign of f’ on either side of a critical point:

• f’ goes from positive to negative: local maximum (the function was rising, now it’s falling)
• f’ goes from negative to positive: local minimum (the function was falling, now it’s rising)
• f’ doesn’t change sign: no extremum (just a flat spot)

At a critical point c where f’(c) = 0:

• f”(c) is positive: local minimum (concave up, like a bowl)
• f”(c) is negative: local maximum (concave down, like a hill)
• f”(c) = 0: inconclusive, use the first derivative test instead

On a closed interval [a, b], the absolute max and min must occur at either a critical point inside the interval or at one of the endpoints. Evaluate f at all candidates and compare.

Example 1: First Derivative Test

Find the local extrema of f(x) = x³ - 3x² - 9x + 5.

$$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)$$

Critical points: x = -1 and x = 3.

Check the sign of f’ in each interval:

• x less than -1: f’(-2) = 3(4 + 4 - 3) = 15, positive (rising)
• Between -1 and 3: f’(0) = 3(0 - 0 - 3) = -9, negative (falling)
• x greater than 3: f’(4) = 3(16 - 8 - 3) = 15, positive (rising)

f’ goes from positive to negative at x = -1: local max. f’ goes from negative to positive at x = 3: local min.

Example 2: Second Derivative Test

For f(x) = x⁴ - 4x², find and classify the critical points.

$$f'(x) = 4x^3 - 8x = 4x(x^2 - 2)$$

Critical points: x = 0, x = sqrt(2), x = -sqrt(2).

$$f''(x) = 12x^2 - 8$$

f”(0) = -8, which is negative: local max at x = 0. f”(sqrt(2)) = 12(2) - 8 = 16, which is positive: local min at x = sqrt(2). f”(-sqrt(2)) = 16, positive: local min at x = -sqrt(2).

Example 3: Absolute extrema on [0, 5]

Find the absolute max and min of f(x) = x³ - 6x² + 9x + 2 on [0, 5].

$$f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$$

Critical points in [0, 5]: x = 1 and x = 3.

Evaluate f at all candidates:

• f(0) = 2
• f(1) = 1 - 6 + 9 + 2 = 6
• f(3) = 27 - 54 + 27 + 2 = 2
• f(5) = 125 - 150 + 45 + 2 = 22

f(x) = x³ - 6x² + 9x + 2 local max at (1, 6) local min at (3, 2) abs max at (5, 22)

The absolute maximum is 22 at x = 5 (the right endpoint). The absolute minimum is 2, occurring at both x = 0 and x = 3. The local max at (1, 6) is not the absolute max because the endpoint at x = 5 is higher. This is why you always check endpoints on closed intervals.

Finding maxima and minima is the core of optimization:

• Business: maximize profit or minimize cost given constraints
• Engineering: find the dimensions that maximize strength while minimizing material
• Physics: find the angle that maximizes projectile range, or the path that minimizes travel time
• Medicine: find the dosage that maximizes effectiveness while minimizing side effects