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Class 12 Statistics Notes · GSEB

Increasing and Decreasing

Increasing and Decreasing — connect the sign of f' to rising and falling behaviour with interactive sign charts and solved point tests. GSEB Class 12 Commerce Statistics notes.

Last updated: 24 Sep 2026

Notes

Increasing and Decreasing Functions

Increasing at x = a

For a small positive h: f(a + h) > f(a) and f(a) > f(a − h) — the function climbs through a.

Derivative test

f′(a)>0f'(a) > 0

Decreasing at x = a

For a small positive h: f(a + h) < f(a) and f(a) < f(a − h) — the function falls through a.

Derivative test

f′(a)<0f'(a) < 0

One-sentence VSQ answers

Increasing at a ⟺ f′(a) > 0. Decreasing at a ⟺ f′(a) < 0. Stationary at a ⟺ f′(a) = 0. The sign of the first derivative is the entire story.

Stationary Points

Stationary Point
A point where the first order derivative equals zero — f′(a) = 0. These are the only places where a maximum or minimum can occur, and the borders of every sign chart.

f′ > 0

↗ rising — increasing

f′ = 0

— flat — stationary (peak? valley?)

f′ < 0

↘ falling — decreasing

Exam tip — the four-step sign chart

1) Find f′(x). 2) Solve f′(x) = 0 for critical points. 3) Test one sample value in each interval. 4) Read the sign → + means increasing, − means decreasing. This exact procedure appears in Section D and F questions.

Explore: Sign Chart on the Graph

Green segments rise, red segments fall, amber dashed lines mark the stationary points. Drag x, click an interval of the sign chart — or type your own f(x) to rebuild everything automatically.

Sign Chart Lab — f(x) = x³ − 3x² + 7, f′(x) = 3x² − 6x
Try your own function (polynomials only)

The lab differentiates your f(x), solves f′ = 0, splits the curve into rising (green) and falling (red) segments, and builds the sign chart automatically.

02f′ = 9↗ increasing (green)↘ decreasing (red)

f(x)

3

f′(x)

9

Verdict

↗ Increasing (f′ > 0)

Sign chart — click an interval

Check at x = −1: f′(−1) = 9 → positive → function is increasing on x < 0.

Critical points split the map

f′(x) = 3x² − 6x = 0 at x = 0 and x = 2. These stationary points are the borders of the sign chart — between them the derivative cannot change sign without passing through zero.

Test at Specific Points — Worked Examples

Solved Example

Problem

If f(x) = x² − 4x, decide whether f is increasing or decreasing at x = −1, x = 0 and x = 3.

Solution

Decreasing at x = −1 and x = 0; increasing at x = 3 (stationary point sits at x = 2)

Solved Example

Problem

Decide whether y = x³ − 3x² + 7 is increasing or decreasing at x = 1 and x = 3.

Solution

Decreasing at x = 1, increasing at x = 3

Solved Example

Problem

Is y = 12 + 4x − 7x² increasing or decreasing at x = 2? (Section C)

Solution

Decreasing at x = 2 (dy/dx = −24)

Quick Reference

First derivative sign → behaviour
Sign of f′BehaviourArrowWhat to do next
f′(x) > 0Increasing↗function climbing — safe zone for growth
f′(x) < 0Decreasing↘function falling — costs rising or sales dropping
f′(x) = 0Stationary—candidate peak/valley → apply the f″ test

Negative example — skipping the sign chart

A student solves f′(x) = 0, finds x = 2, and declares “maximum at 2” without checking signs. For f(x) = x³ − 3x² + 7, x = 2 is actually a minimum(f′ goes − to +). f′ = 0 only means “flat” — the sign change around it decides the verdict.

Key Takeaways

Key Takeaways

  • ★ f′(a) &gt; 0 → increasing · f′(a) &lt; 0 → decreasing · f′(a) = 0 → stationary.
  • Stationary points (f′ = 0) are where maxima/minima may live — and the borders of every sign chart.
  • Sign chart procedure: f′ → solve f′ = 0 → test a sample in each interval → read +/−.
  • A single point test is not enough to classify a peak — you need the sign change around the critical point.
  • So what? Finding where profit rises and where it falls — before choosing the optimal output — is exactly this skill with ₹ attached.

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