Local maxima and minima: Class 12 practice

What you should know first

• Differentiation

The idea behind local maxima and minima

At a stationary point f′(x) = 0. If f″(x) < 0 there it is a local maximum; if f″(x) > 0 a local minimum.

Step-by-step method

1. Solve f′(x) = 0.
2. Find f″(x) at each point.
3. Classify and find the function value.

Worked example 1 (Basic)

Find the minimum value of f(x) = x² + 2x + 3.

1. f′ = 2x + 2 = 0 → x = -1
2. f″ = 2 > 0 → minimum
3. f(-1) = 2
4. Answer: 2

Worked example 2 (Standard)

Find the local maximum and minimum values of f(x) = x³ − 3x − 3.

1. f′ = 3x² − 3 = 0 → x = ±1
2. f″ = 6x: x = −1 max, x = 1 min
3. Max -1, min -5
4. Answer: Max -1 at x = −1; min -5 at x = 1

Worked example 3 (Challenge)

Find the local maximum and minimum values of f(x) = x³ − 12x − 3.

1. f′ = 3x² − 12 = 0 → x = ±2
2. f″ = 6x: x = −2 max, x = 2 min
3. Max 13, min -19
4. Answer: Max 13 at x = −2; min -19 at x = 2

Common mistakes and misconceptions

• Forgetting to substitute back into f
• Mixing up the signs of f″

Practice questions

1. (Basic) Find the minimum value of f(x) = x² + 4x − 3.
2. (Basic) Find the minimum value of f(x) = x² + 6x + 6.
3. (Standard) Find the local maximum and minimum values of f(x) = x³ − 3x + 4.
4. (Standard) Find the local maximum and minimum values of f(x) = x³ − 48x − 1.
5. (Challenge) Find the local maximum and minimum values of f(x) = x³ − 48x + 0.
6. (Challenge) Find the local maximum and minimum values of f(x) = x³ − 48x − 7.

Answer key with full working

• Q1: f′ = 2x + 4 = 0 → x = -2 → f″ = 2 > 0 → minimum → f(-2) = -7 → Answer: -7
• Q2: f′ = 2x + 6 = 0 → x = -3 → f″ = 2 > 0 → minimum → f(-3) = -3 → Answer: -3
• Q3: f′ = 3x² − 3 = 0 → x = ±1 → f″ = 6x: x = −1 max, x = 1 min → Max 6, min 2 → Answer: Max 6 at x = −1; min 2 at x = 1
• Q4: f′ = 3x² − 48 = 0 → x = ±4 → f″ = 6x: x = −4 max, x = 4 min → Max 127, min -129 → Answer: Max 127 at x = −4; min -129 at x = 4
• Q5: f′ = 3x² − 48 = 0 → x = ±4 → f″ = 6x: x = −4 max, x = 4 min → Max 128, min -128 → Answer: Max 128 at x = −4; min -128 at x = 4
• Q6: f′ = 3x² − 48 = 0 → x = ±4 → f″ = 6x: x = −4 max, x = 4 min → Max 121, min -135 → Answer: Max 121 at x = −4; min -135 at x = 4

Syllabus note

This skill is listed under "Application of Derivatives" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

• Application of Derivatives chapter notes (Class 12)
• Increasing and decreasing intervals
• Rates of change
• Tangents and normals
• Derivatives of polynomials and slope of a tangent
• Definite integrals of polynomials
• Evaluating 2×2 and 3×3 determinants
• Matrix multiplication
• All Class 12 practice

Frequently asked questions

Which chapter is local maxima and minima in?

"Application of Derivatives", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.