chapter

Maxima, Minima and Applications of Derivatives Notes for GATE DA

Maxima, Minima and Applications of Derivatives notes for GATE DA: 18 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice q

maxima minima and applications of derivatives notes

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey @keyframes fadeInUp { from { opacity: 0; transform: translateY(15px); } to { opacity: 1; transform: translateY(0); } } .step-item { animation: fadeInUp 0.6s ease-out forwards; opacity: 0; } .step-1 { animation-delay: 0.1s; } .step-2 { animation-delay: 0.4s; } 01 Critical Points & Second Derivative Test Current Topic • Foundation 02 Polynomial Extrema & Interval Analysis Global Max/Min, Boundary Checks Goal for this topic: Master the identification of critical points and use the second derivative to classify them as local maxima, minima, or saddle points.

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots A critical point of a function f (x) occurs at x = c if: f^{'} (c) = 0 (The tangent is horizontal) OR f^{'} (c) does not exist (Sharp corner or vertical tangent) f^{'} (c) = 0 or f^{'} (c) is undefined Why care? Local maxima (peaks) and local minima (valleys) can only occur at critical points. If the slope is not zero and exists, you are still going up or down.

Visualizing Concavity and Extrema

Geometric Interpretation f''(c) > 0 Local Min Concave Up (\cup) f''(c) < 0 Local Max Concave Down (\cap) Geometric intuition is often faster than calculating in multiple-choice questions.

15 more cards in this chapter

Try a question

Answer it here to see how it works. Nothing is recorded until you sign in.

Question 1

Level 1: Warm-up

Let f (x) = e^{x} - 2 x, which has a critical point at x = ln 2 . Using the Second Derivative Test, what does the test determine about x = ln 2 ?

Question 2

Level 1: Warm-up

Let f (x) = - x^{2} + 6 x - 5 . The function has a critical point at x = 3 . What does the Second Derivative Test conclude about x = 3 ?

Question 3

Level 1: Warm-up

For f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1, consider the assertion: " x = 3 is a local minimum" and the reason: " f^{''} (3) = 12 > 0 ".

Question 4

Level 1: Warm-up

For f (x) = x^{3} - 3 x, consider the two statements about x = 1 : (I) f^{'} (1) = 0 . (II) x = 1 is a critical point of f . Which conclusion is correct?

Question 5

Level 1: Warm-up

Consider f (x) = x^{1/3} . Which statement correctly classifies x = 0 ?

Question 6

Level 1: Warm-up

How many critical points does the function f (x) = ∣ x^{2} - 4∣ have in the set of real numbers?

Question 7

Level 1: Warm-up

How many distinct critical points does the function f (x) = x^{4} - 4 x^{3} + 6 x^{2} have in the set of real numbers?

Question 8

Level 1: Warm-up

Given f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1 with f^{''} (x) = 3 x^{2} - 4 x - 3, construct the value of f^{''} (- 1) and classify x = - 1 .

Question 9

Level 1: Warm-up

Given f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1 with f^{''} (x) = 3 x^{2} - 4 x - 3, construct the value of f^{''} (0) and classify x = 0 .

Question 10

Level 1: Warm-up

Let f (x) = - cos x . At x = π, the function has a critical point. What does the Second Derivative Test conclude about x = π ?

Free preview ends here

Login to view the complete notes

Creating an account is free. You get the rest of this chapter, step-by-step solutions, and a study plan built around the topics you are actually weak at.

Why MastersUp

Personalised first. High quality throughout.

Most platforms hand everyone the same content. Here the content moves with your performance, topic by topic.

Built around you, not around a syllabus PDF

Every answer you give moves your topic-level intelligence rate. The next question, the next revision card and tomorrow's plan all change with it.

Revision that hits your weak spots

We only revise topics you have actually attempted and are still below the safe bar on — never the same chapter on repeat.

Questions calibrated to the real exam

Each question carries a measured toughness. You are served a rung above your current level, so practice keeps stretching you.

Notes written for recall, not for volume

Full lesson cards for first study, curated short-note cards for the last mile — with derivations, traps and exam patterns marked.

One place for everything

Notes, chapter practice, previous-year questions, test series and full-length papers — all feeding one picture of your preparation.

Honest progress

No vanity streaks. Progress here means chapters mastered and accuracy that held up on harder questions.

Unlock the whole course

Full notes and short notes, the complete question bank with worked solutions, mock tests, full-length papers, and an adaptive plan that rebuilds itself as you improve.

View all plans Start free

Maxima, Minima and Applications of Derivatives Notes for GATE DA

Maxima, Minima and Applications of Derivatives notes for GATE DA: 18 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey

Critical Points & Second Derivative Test

Current Topic • Foundation

Polynomial Extrema & Interval Analysis

Global Max/Min, Boundary Checks

Goal for this topic: Master the identification of critical points and use the second derivative to classify them as local maxima, minima, or saddle points.

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots

A critical point of a function $f (x)$ occurs at $x = c$ if:

$f^{'} (c) = 0$ (The tangent is horizontal)
OR $f^{'} (c)$ does not exist (Sharp corner or vertical tangent)

f^{'} (c) = 0 or f^{'} (c) is undefined

Why care?
Local maxima (peaks) and local minima (valleys) can only occur at critical points. If the slope is not zero and exists, you are still going up or down.

Visualizing Concavity and Extrema

Geometric Interpretation

Concave Up (

\cup

)

Concave Down (

\cap

)

Geometric intuition is often faster than calculating in multiple-choice questions.

Step-by-Step Method for Classification

The Algorithm

Differentiate: Find $f^{'} (x)$ .
Find Critical Points: Solve $f^{'} (x) = 0$ for $x$ . Let solutions be $c_{1}, c_{2}, \dots$ .
Second Derivative: Find $f^{''} (x)$ .
Evaluate: Substitute each $c_{i}$ into $f^{''} (x)$ .

Result at $c_{i}$	Conclusion
$f^{''} (c_{i}) > 0$	Local Minimum at $c_{i}$
$f^{''} (c_{i}) < 0$	Local Maximum at $c_{i}$
$f^{''} (c_{i}) = 0$	Test Fails (Use First Derivative Test)

Maxima, Minima and Applications of Derivatives: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Calculus and Optimization MCQ

Let $f (x) = e^{x} - 2 x$ , which has a critical point at $x = ln 2$ . Using the Second Derivative Test, what does the test determine about $x = ln 2$ ?

A.
Local minimum
B.
Local maximum
C.
Inflection point
D.
Test fails

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test: sign of $f^{''}$ at the critical point classifies it.

Step 1: $f^{'} (x) = e^{x} - 2$ , and indeed $f^{'} (ln 2) = 2 - 2 = 0$ .

Step 2: $f^{''} (x) = e^{x}$ , so $f^{''} (ln 2) = e^{l n 2} = 2 > 0$ .

Step 3: Positive second derivative ⇒ concave up ⇒ local minimum by the test.

Note: the question asks only what the TEST determines; no boundary/global claim is needed.

Answer: A

Question 2 · Calculus and Optimization MCQ

Let $f (x) = - x^{2} + 6 x - 5$ . The function has a critical point at $x = 3$ . What does the Second Derivative Test conclude about $x = 3$ ?

A.
Local minimum
B.
Inflection point
C.
Local maximum
D.
Test is inconclusive

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test — sign of $f^{''}$ at the critical point classifies it.

Step 1: Compute $f^{'} (x) = - 2 x + 6$ . Verify: $f^{'} (3) = - 6 + 6 = 0$ . Confirmed critical point.

Step 2: Compute $f^{''} (x) = - 2$ .

Step 3: Evaluate: $f^{''} (3) = - 2 < 0$ .

Step 4: Negative second derivative ⇒ concave down ⇒ local maximum by the test.

Answer: C

Question 3 · Calculus and Optimization MCQ

For $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ , consider the assertion: " $x = 3$ is a local minimum" and the reason: " $f^{''} (3) = 12 > 0$ ".

A.
Both assertion and reason are true, and the reason is the correct explanation
B.
Both are true, but the reason is not the correct explanation
C.
Assertion is true, reason is false
D.
Assertion is false, reason is true

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution to verify both the assertion and the reason, then check their logical link.

Step 1: Recall $f^{'} (x) = x^{3} - 2 x^{2} - 3 x$ and $f^{''} (x) = 3 x^{2} - 4 x - 3$ .

Step 2: Verify the reason: $f^{''} (3) = 3 (3)^{2} - 4 (3) - 3 = 27 - 12 - 3 = 12$ . Since $12 > 0$ , the reason is TRUE.

Step 3: Verify the assertion: Since $f^{'} (3) = 3^{3} - 2 (3)^{2} - 3 (3) = 27 - 18 - 9 = 0$ , $x = 3$ is a critical point. And since $f^{''} (3) = 12 > 0$ , the Second Derivative Test confirms it is a local minimum. The assertion is TRUE.

Step 4: Check the link: The reason ( $f^{''} (3) > 0$ ) is the exact condition of the Second Derivative Test that proves the assertion (local minimum). Thus, the reason is the correct explanation.

Answer: A

Question 4 · Calculus and Optimization MCQ

For $f (x) = x^{3} - 3 x$ , consider the two statements about $x = 1$ :

(I) $f^{'} (1) = 0$ .

(II) $x = 1$ is a critical point of $f$ .

Which conclusion is correct?

A.
Both I and II are true
B.
I is true, II is false
C.
I is false, II is true
D.
Both I and II are false

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the definition of a critical point ( $f^{'} (c) = 0$ or $f^{'} (c)$ undefined).

Step 1: Compute $f^{'} (x) = 3 x^{2} - 3$ .

Step 2: Evaluate at $x = 1$ : $f^{'} (1) = 3 (1)^{2} - 3 = 0$ . So statement I is TRUE.

Step 3: Since $f^{'} (1) = 0$ (and the derivative exists everywhere for a polynomial), $x = 1$ satisfies the definition of a critical point. Statement II is TRUE.

Step 4: Both statements hold → option A.

Answer: A

Question 5 · Calculus and Optimization MCQ

Consider $f (x) = x^{1/3}$ . Which statement correctly classifies $x = 0$ ?

A.
It is a critical point because $f^{'} (0)$ does not exist
B.
It is a critical point because $f^{'} (0) = 0$
C.
It is not a critical point because $f (0) = 0$
D.
It is not a critical point because $f$ is continuous at $0$

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the critical-point definition — two branches: $f^{'} (c) = 0$ OR $f^{'} (c)$ undefined.

Step 1: Compute $f^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}}$ .

Step 2: At $x = 0$ , the denominator is zero, so $f^{'} (0)$ does not exist (vertical tangent).

Step 3: Since $f^{'} (0)$ is undefined, $x = 0$ satisfies the second branch of the critical-point definition. It is a critical point.

Step 4: Option A states this correctly. Option B incorrectly claims $f^{'} (0) = 0$ . Options C and D confuse function value or continuity with the derivative condition.

Answer: A

Question 6 · Calculus and Optimization MCQ

How many critical points does the function $f (x) = ∣ x^{2} - 4∣$ have in the set of real numbers?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the critical-point definition across the pieces of an absolute-value function.

Step 1: The expression inside the absolute value is zero at $x = 2$ and $x = - 2$ . At these points, the function has sharp corners, so $f^{'} (x)$ does not exist. These are two critical points.

Step 2: For $- 2 < x < 2$ , $f (x) = 4 - x^{2}$ , so $f^{'} (x) = - 2 x$ . Setting $f^{'} (x) = 0$ gives $x = 0$ . This is a third critical point.

Step 3: For $x > 2$ and $x < - 2$ , $f (x) = x^{2} - 4$ , so $f^{'} (x) = 2 x$ . Setting $f^{'} (x) = 0$ gives $x = 0$ , which is not in these intervals. No additional critical points.

Step 4: Total critical points: $x = - 2, 0, 2$ . The count is 3.

Answer: C

Question 7 · Calculus and Optimization MCQ

How many distinct critical points does the function $f (x) = x^{4} - 4 x^{3} + 6 x^{2}$ have in the set of real numbers?

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: Critical points occur where $f^{'} (x) = 0$ . For polynomials, we find the distinct real roots of the derivative.

Step 1: Find the first derivative.

$f^{'} (x) = 4 x^{3} - 12 x^{2} + 12 x$ .

Step 2: Factor out the greatest common factor.

$f^{'} (x) = 4 x (x^{2} - 3 x + 3)$ .

Step 3: Set each factor to zero and solve for real roots.

$4 x = 0 ⟹ x = 0$ .

For the quadratic $x^{2} - 3 x + 3 = 0$ , check the discriminant:

$Δ = b^{2} - 4 a c = (- 3)^{2} - 4 (1) (3) = 9 - 12 = - 3$ .

Since $Δ < 0$ , the quadratic has no real roots.

Step 4: Count the distinct real roots.

The only real root is $x = 0$ .

Answer: There is exactly 1 distinct critical point.

Question 8 · Calculus and Optimization MCQ

Given $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ with $f^{''} (x) = 3 x^{2} - 4 x - 3$ , construct the value of $f^{''} (- 1)$ and classify $x = - 1$ .

A.
$f^{''} (- 1) = 4 > 0$ , local minimum
B.
$f^{''} (- 1) = - 4 < 0$ , local maximum
C.
$f^{''} (- 1) = 2 > 0$ , local minimum
D.
$f^{''} (- 1) = 0$ , test fails

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution — build $f^{''} (- 1)$ term by term from the supplied formula, keeping signs intact.

Step 1: $f^{''} (- 1) = 3 (- 1)^{2} - 4 (- 1) - 3$ .

Step 2: Term values: $3 (1) = 3$ ; $- 4 (- 1) = + 4$ ; constant $- 3$ .

Step 3: Sum: $3 + 4 - 3 = 4 > 0$ ⇒ concave up ⇒ local minimum at $x = - 1$ .

Step 4: Only option A reports both the correct constructed value and the correct classification.

Answer: A

Question 9 · Calculus and Optimization MCQ

Given $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ with $f^{''} (x) = 3 x^{2} - 4 x - 3$ , construct the value of $f^{''} (0)$ and classify $x = 0$ .

A.
$f^{''} (0) = 3 > 0$ , local minimum
B.
$f^{''} (0) = 0$ , test fails
C.
$f^{''} (0) = - 3 < 0$ , local minimum
D.
$f^{''} (0) = - 3 < 0$ , local maximum

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution — build $f^{''} (0)$ term by term from the supplied formula.

Step 1: $f^{''} (0) = 3 (0)^{2} - 4 (0) - 3$ .

Step 2: Term values: $3 (0) = 0$ ; $- 4 (0) = 0$ ; constant $- 3$ .

Step 3: Sum: $0 + 0 - 3 = - 3 < 0$ .

Step 4: Negative second derivative ⇒ concave down ⇒ local maximum at $x = 0$ .

Step 5: Only option D reports both the correct constructed value and the correct classification.

Answer: D

Question 10 · Calculus and Optimization MCQ

Let $f (x) = - cos x$ . At $x = π$ , the function has a critical point. What does the Second Derivative Test conclude about $x = π$ ?

A.
Local minimum
B.
Local maximum
C.
Inflection point
D.
Test fails

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test to a trigonometric function.

Step 1: Compute $f^{'} (x) = sin x$ . Verify: $f^{'} (π) = sin (π) = 0$ . Confirmed critical point.

Step 2: Compute $f^{''} (x) = cos x$ .

Step 3: Evaluate: $f^{''} (π) = cos (π) = - 1$ .

Step 4: Since $f^{''} (π) = - 1 < 0$ , the function is concave down at $x = π$ . By the Second Derivative Test, this indicates a local maximum.

Answer: B

maxima minima and applications of derivatives notes

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots

Visualizing Concavity and Extrema

Geometric Interpretation

Try a question

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Maxima, Minima and Applications of Derivatives Notes for GATE DA

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots

Visualizing Concavity and Extrema

Geometric Interpretation

Step-by-Step Method for Classification

The Algorithm

Maxima, Minima and Applications of Derivatives: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit