Increasing and decreasing functions

On an interval I, f is increasing if x₁ < x₂ in I gives f(x₁) ≤ f(x₂), and strictly increasing if it gives f(x₁) < f(x₂). Decreasing and strictly decreasing are defined with the inequalities reversed.

A constant function satisfies both ≤ and ≥, so it counts as increasing and as decreasing on any interval, but it is strictly neither.

Derivative test: suppose f is differentiable on (a, b) and continuous on the closed interval [a, b]. If f′ > 0 on (a, b), f is increasing on [a, b]; if f′ < 0, f is decreasing; if f′ = 0 throughout, f is constant.

The proof uses the mean value theorem: f(x₂) − f(x₁) = f′(c)(x₂ − x₁) for some c between them, so the sign of f′ fixes the sign of the change.

Worked example: f(x) = 7x − 3 has f′ = 7 > 0, so it is increasing on R.

Worked example: f(x) = x³ − 3x² + 4x has f′ = 3x² − 6x + 4 = 3(x − 1)² + 1, which is always positive, so f is increasing on R.

Worked example: cos x has derivative −sin x, which is negative on (0, π) and positive on (π, 2π). So cos x decreases on (0, π) and increases on (π, 2π), and is neither on the whole of (0, 2π).

f′ may vanish at single points and f can still be strictly increasing: x³ has f′(0) = 0 yet rises through 0.