Increasing and Decreasing Functions: Test and Examples

A function is increasing on an interval when its graph rises as you move left to right, and decreasing when it falls. The derivative gives a precise test for this: where f'(x) is positive the function climbs, and where f'(x) is negative it descends. This turns a visual idea into something you can prove with algebra.

The test matters because monotonicity — being consistently increasing or decreasing — underlies almost everything in curve analysis. It locates local maxima and minima, justifies inequalities, and tells you whether a model (like a growing population or a cooling object) behaves the way you expect on each part of its domain.

Key Concepts

• Increasing function: f is increasing on an interval if x₁ smaller than x₂ implies f(x₁) smaller than f(x₂) for all x₁, x₂ in the interval.
• Decreasing function: f is decreasing on an interval if x₁ smaller than x₂ implies f(x₁) greater than f(x₂).
• First derivative test for monotonicity: if f'(x) greater than 0 on an interval, f is increasing there; if f'(x) smaller than 0, f is decreasing there.
• Critical points: points where f'(x) = 0 or f' is undefined split the domain into intervals where the sign of f' cannot change.

How It Works

1. Step 1: Compute f'(x).
2. Step 2: Find the critical points by solving f'(x) = 0 (and noting where f' is undefined).
3. Step 3: Mark the critical points on a number line to divide the domain into intervals.
4. Step 4: Pick a test value inside each interval and check the sign of f' there.
5. Step 5: Conclude: positive sign means increasing, negative sign means decreasing on that interval.

Worked Example

Find the intervals where f(x) = 2x³ − 3x² − 12x + 5 is increasing or decreasing.

1. Step 1: Differentiate: f'(x) = 6x² − 6x − 12.
2. Step 2: Factor: 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1). Critical points are x = −1 and x = 2.
3. Step 3: The intervals are (−∞, −1), (−1, 2), and (2, ∞).
4. Step 4: Test signs: at x = −2, f' = 6(−4)(−1) = 24, positive. At x = 0, f' = 6(−2)(1) = −12, negative. At x = 3, f' = 6(1)(4) = 24, positive.
5. Step 5: f is increasing on (−∞, −1) and (2, ∞), and decreasing on (−1, 2).

Common Misconceptions

• "f'(x) = 0 at a point means the function is flat there." Only instantaneously — the function can still be strictly increasing through that point, as with y = x³ at x = 0.
• "A positive derivative at one point means increasing everywhere." Monotonicity is an interval property; the sign of f' must hold across the whole interval, which is why you test each interval separately.
• "Strictly increasing means the derivative is never zero." Wrong — x³ is strictly increasing everywhere even though its derivative is zero at x = 0.

Real-World Applications

• Economics: checking where a demand or cost function is increasing or decreasing tells you the price ranges in which raising prices helps or hurts revenue.
• Medicine: a drug concentration curve that is increasing means the drug is still being absorbed; decreasing means it is being eliminated.
• Physics: velocity increasing means speeding up in the positive direction — the sign of acceleration (the derivative of velocity) determines this.
• Machine learning: gradient descent relies on the function decreasing along the search direction, which is exactly a monotonicity condition.

To recap: find where f' = 0, split the domain into intervals, test the sign of f' in each, and read off increasing where the sign is positive and decreasing where it is negative.