Maxima, minima and monotonicity

Monotonicity

A function f is increasing on an interval if f′(x) > 0 there and decreasing if f′(x) < 0 there. To find these intervals: 1. Find f′(x) and solve f′(x) = 0 to get the critical points. 2. Split the number line at these points and check the sign of f′(x) on each piece.

Example: f(x) = x² − 4x + 1 has f′(x) = 2x − 4, which is negative for x < 2 and positive for x > 2. So f decreases on (−∞, 2) and increases on (2, ∞).

Local maxima and minima

• First derivative test: if f′ changes from + to − at c, f has a local maximum at c; from − to +, a local minimum.
• Second derivative test: if f′(c) = 0 and f″(c) < 0, local maximum; if f″(c) > 0, local minimum; if f″(c) = 0, the test fails.

Example: f(x) = 2x³ − 6x + 5 has f′(x) = 6x² − 6 = 0 at x = ±1. f″(x) = 12x, so f″(1) = 12 > 0 and x = 1 is a local minimum with value 1.

Absolute extrema on a closed interval

On [a, b], the absolute maximum and minimum of a continuous function occur at critical points inside the interval or at the end points. Evaluate f at all of them and compare. For f(x) = x³ − 12x on [−3, 3]: f(−3) = 9, f(−2) = 16, f(2) = −16, f(3) = −9. The absolute maximum is 16 and the absolute minimum is −16.

Applied problems

1. Express the quantity to be optimised in terms of one variable, using the given condition. 2. Differentiate, set the derivative to zero and solve. 3. Confirm with the second derivative and answer in context. Example: 100 m of fencing encloses a rectangular field on three sides, with a river on the fourth. If each side perpendicular to the river is y, the area is A = y(100 − 2y). A′ = 100 − 4y = 0 gives y = 25, so the maximum area is 25 × 50 = 1250 m².

Key points

• f′ > 0: increasing; f′ < 0: decreasing.
• Local extremum test: sign change of f′, or sign of f″.
• Always check end points for absolute extrema.
• A local maximum need not be the absolute maximum.

Common mistakes

• Ignoring end points on a closed interval.
• Reporting the x-value when the question asks for the maximum value.