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Class 12 Statistics Notes · GSEB

Maxima and Minima

Maxima and Minima — apply the f' = 0 and f' test to find peaks, valleys and function values. GSEB Class 12 Commerce Statistics notes with a step-by-step finder.

Last updated: 24 Sep 2026

Notes

Maximum and Minimum Values

Maximum at x = a

f(a) > f(a + h) and f(a) > f(a − h) for small positive h — the peak of a local hill.

Conditions

f′(a)=0 and f′′(a)<0f'(a) = 0 \ \text{and}\ f''(a) < 0

Minimum at x = a

f(a) < f(a + h) and f(a) < f(a − h) for small positive h — the bottom of a local valley.

Conditions

f′(a)=0 and f′′(a)>0f'(a) = 0 \ \text{and}\ f''(a) > 0

Maximum ≠ largest value

A function can have many maxima (local hills) but only one largestvalue (global). “Maximum at x = a” only claims the function is highest in a small neighbourhood of a.

The Four-Step Method

1

Differentiate

Find f′(x) of the given function.

2

Set f′(x) = 0 and solve

The roots are the stationary points (candidates only).

3

Apply the f″ test

f″ < 0 at a root → maximum · f″ > 0 → minimum · f″ = 0 → inconclusive, use sign change of f′.

4

Plug back into f(x)

Substitute each x into the ORIGINAL function to get the actual max/min value.

Memory hook

f′ finds where · f″ decides which · f(x) tells you how much. Where → which → how much. Never skip step 4 — marks live there.

Explore: Peak and Valley Finder

Perform the same four steps you would write in the exam: differentiate → solve f′ = 0 → apply the f″ test → plug back into f(x). Each click unlocks the next step on the curve and in the cards.

Peak & Valley Finder — f(x) = 2x³ + 3x² − 12x − 4
Try your own function (polynomials only)

Type powers with ^ — like 2x^3 + 3x^2 - 12x - 4. The solver differentiates, finds f′ = 0, applies the f″ test and graphs it — the same four steps.

Work through steps 1–4 in order — the same four steps you will write in the exam…

Condition Reference

Second order derivative test for stationary points
ConditionResultShape
f′ = 0, f″ < 0Maximumpeak (∩)
f′ = 0, f″ > 0Minimumvalley (∪)
f′ = 0, f″ = 0Inconclusivetest the sign of f′ around the point
Value at the max/minSubstitute x into f(x)never leave it at x alone

Solved Examples

Solved Example

Problem

Find the maximum and minimum values of f(x) = 2x³ + 3x² − 12x − 4.

Solution

Maximum value = 16 at x = −2; Minimum value = −11 at x = 1

Solved Example

Problem

Find the maximum and minimum values of y = x³ − 2x² − 4x − 1.

Solution

Maximum value = 13/27 at x = −2/3; Minimum value = −9 at x = 2

Solved Example

Problem

Find values of x that max/min y = 2x³ − 15x² + 36x + 12. (Section F)

Solution

Maximum 40 at x = 2; Minimum 39 at x = 3

Real Business Stakes

Find the valley — minimise cost

A steel plant with C = 10x² − 1000x + 50000 finds C′ = 0 at x = 50, and C″ = 20 > 0 → minimum. Producing 50 tons costs ₹25,000 — cheaper than 40 or 60.

Find the peak — maximise profit

Zomato-style dashboards optimise delivery incentive spend: too little incentive → orders drop, too much → margin erodes. The profitable middle is the maximum of the profit function — f′ = 0, f″ < 0.

Negative example — stopping at step 2

Solving f′ = 0 and writing “maximum at x = −2” without f″ or without the value f(−2) = 16 loses marks in Section E/F. Examiners want: conditions applied and the final value.

Key Takeaways

Key Takeaways

  • ★ Maximum: f′(a) = 0 AND f″(a) &lt; 0. Minimum: f′(a) = 0 AND f″(a) &gt; 0.
  • Stationary points are only candidates — the f″ test (or sign change of f′) delivers the verdict.
  • Always substitute back into f(x) to get the actual maximum/minimum VALUE, not just the x.
  • Maximum means locally highest in a neighbourhood — not necessarily the largest value over the whole domain.
  • Procedure: differentiate → set to zero → f″ test → substitute. Where → which → how much.
  • So what? Cheapest production scale and best production scale for profit are literally the valley and peak of cost and profit curves.

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