A derivative suggests where to look. The domain and a comparison decide what is truly largest or smallest.
JEE Main · JEE Advanced55 min study guide9 indexed sections
BY THE END, YOU CAN
Distinguish local extrema from absolute extrema and values from locations.
Build a complete candidate list including endpoints and nondifferentiable points.
Use derivative sign changes and curvature with their correct hypotheses.
Translate an optimization story into a one-variable function on a justified domain.
BEFORE YOU START
Function domains, interval notation and solving inequalities.
Derivatives of polynomial, rational, exponential and basic trigonometric functions.
Product, quotient and chain rules.
Continuity and one-sided limits; basic sign charts.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Main 2025. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Maxima and Minima.
JEE Main · 2025 · 2 April · Shift 1 · Paper 1 · Mathematics · Q8 · ID 603421908
Paraphrased exam question
Answer checked against the official final key. Worked solution written for this lesson.
For a > 0, let f(x) = 2x³ − 9ax² + 12a²x + 1. Its local maximum occurs at x = p and local minimum at x = q. Given p² = q, calculate f(3).
Reveal answer & worked solution
Answer: f(3) = 37 (official option ID 6034213089).
Source: JEE (Main) 2025, 2 April, first shift, Paper 1, question ID 603421908; official final key, 18 April 2025. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. What does “best” mean?
A rectangle can become wider while its height shrinks. A function can rise for a while and then fall. Optimization asks which admissible choice produces the largest or smallest result. The word admissible matters: a mathematically convenient x may describe a negative length, a forbidden denominator, or a point outside the interval in the question.
Think of a graph as a landscape. A local maximum is higher than nearby points in the domain; an absolute maximum is at least as high as every allowed point. The tallest hill in one neighbourhood may still lie below a distant peak. A local minimum has the corresponding nearby comparison. Neither comparison requires the value to be positive.
There are two different answers to report. The location x = c tells you where an extreme occurs. The value f(c) tells you how high or low it is. In an applied problem, the question might instead want a width, an area, or both. Label these quantities rather than returning a bare number.
We discuss two-sided local extrema at interior points unless explicitly saying “one-sided endpoint extremum.” This avoids a common convention mismatch: some definitions call an endpoint a local extremum relative to the domain. Regardless of terminology, endpoints must always be considered for absolute extrema when they belong to the domain.
Absolute maximum at c: f(c) ≥ f(x) for every x in the domain
For a local maximum, require the inequality only for domain points sufficiently close to c. Reverse the inequality for a minimum.
02 / UNDERSTAND & APPLY
2. Check whether a best value is attained
A continuous function on a closed, bounded interval [a,b] attains an absolute maximum and an absolute minimum. This is the extreme value theorem. It guarantees existence, but does not tell you where the extrema are or whether they are unique. A constant function attains both everywhere.
Every hypothesis has a job. For f(x) = x on (0,1), values approach 1 but never equal it; there is no maximum. For f(x) = x on [0,1], the endpoint 1 is included and the maximum is 1. For f(x) = x² on the entire real line, a minimum exists but no largest value exists. Removing a hypothesis removes the guarantee, not necessarily the extrema themselves.
Discontinuity can also prevent attainment on a closed interval. Define f(x) = x for 0 ≤ x < 1 and f(1) = 0. Its domain is [0,1], yet there is no largest value: any value below 1 from the rising branch can be exceeded. The missing continuity at 1 is decisive.
An upper bound is not automatically a maximum. The supremum is the least upper bound; it need not be achieved by an allowed input. In exam working, the phrase “approaches 1” must not silently become “maximum 1.” Check whether the candidate belongs to the domain.
ORIGINAL WORKED EXAMPLE
Worked example 1 · Foundation
Find the absolute extrema of f(x) = (x − 2)² on [0,3]. Then answer on the open interval (0,3).
The square is nonnegative and equals zero at x = 2, which belongs to both domains. Thus both have minimum 0.
On [0,3], compare the vertex and endpoints: f(0) = 4, f(2) = 0, f(3) = 1. The maximum is 4 at x = 0.
On (0,3), x = 0 is forbidden. Values approach 4 as x approaches 0 from the right, but no allowed point attains 4.
Every attained value can be exceeded by choosing a sufficiently small positive x. Therefore there is no maximum on (0,3).
Answer [0,3]: minimum 0 at 2, maximum 4 at 0. (0,3): minimum 0 at 2; no maximum.
03 / UNDERSTAND & APPLY
3. Why zero slope is only a candidate
At a differentiable interior local extremum, the derivative must be zero. To see the idea at a local maximum c, compare f(c+h) with f(c). For positive h the difference quotient is nonpositive; for negative h it is nonnegative. If both sides approach the same derivative, that derivative can only be zero. Differentiability and an interior point are essential.
The implication goes in one direction. For f(x) = x³, f′(0) = 0 but the function increases through zero, so there is no extremum there. A horizontal tangent can be a momentary flattening rather than a turn. Conversely, f(x) = |x| has a minimum at zero but no derivative there.
An interior critical point is a domain point where the derivative is zero or does not exist. Some school texts use narrower terminology, so the safest practice is to spell out your candidate list: stationary points, nondifferentiable domain points, and included endpoints. A place where the function itself is undefined is never an extremum location.
For a continuous function on a closed interval, evaluate f at every candidate and compare the resulting values to find absolute extrema. Derivative tests can classify interior behaviour, but they do not remove the endpoint comparison. If a piecewise function has a jump, inspect its assigned value and nearby values directly; a derivative sign chart across a discontinuity alone can mislead.
Interior local extremum + differentiability ⇒ f′(c) = 0
The reverse implication is false. Endpoints and corners need separate checks.
04 / UNDERSTAND & APPLY
4. A sign change explains the turning
On an interval where a function is differentiable, a positive derivative implies increasing behaviour and a negative derivative implies decreasing behaviour. Split the domain at critical points and discontinuities, then choose a test input in each interval to establish the derivative’s sign. Factorization often makes the sign chart quicker and safer than substituting many decimals.
At a point where the function is continuous, a derivative sign change from positive to negative indicates a local maximum: the graph approaches while rising and leaves while falling. Negative to positive indicates a local minimum. No sign change means the first-derivative test does not produce an extremum there; in the standard monotone-through case, the stationary point is neither.
The diagram uses f(x) = x³ − 3x. Its turning points occur at x = −1 and x = 1. Notice that the maximum is only local: farther to the right, the cubic eventually exceeds it. Similarly, the local minimum is not the smallest value over all real inputs.
Do not confuse the sign of f with the sign of f′. A function can be negative while increasing, as this cubic is between some inputs. Heights answer value questions; slopes answer movement questions.
Local turning points of f(x) = x³ − 3x. There are no global extrema on ℝ.
ORIGINAL WORKED EXAMPLE
Worked example 2 · Main-style
Classify the stationary points of f(x) = x³ − 3x and find its absolute extrema on [−2,3].
f′(x) = 3x² − 3 = 3(x − 1)(x + 1), so stationary points are −1 and 1.
The derivative signs are positive on (−∞,−1), negative on (−1,1), and positive on (1,∞). Thus −1 is a local maximum and 1 is a local minimum.
Evaluate the full closed-interval candidate list: f(−2) = −2, f(−1) = 2, f(1) = −2, f(3) = 18.
The endpoint x = 3 beats the local maximum. The absolute minimum occurs at both an endpoint and an interior point.
Answer Local maximum 2 at −1; local minimum −2 at 1. On [−2,3], absolute maximum 18 at 3 and minimum −2 at −2 and 1.
05 / UNDERSTAND & APPLY
5. Curvature is a shortcut, with a stop sign
If f′(c) = 0 and the function is twice continuously differentiable near c, then f″(c) > 0 gives a strict local minimum and f″(c) < 0 gives a strict local maximum. Positive curvature means the slope is increasing through zero; negative curvature means it is decreasing through zero. This is a local classification, not a claim about the entire domain.
If f″(c) = 0, the second-derivative test is inconclusive. The examples x⁴, −x⁴ and x³ all have first and second derivatives zero at the origin, but give a minimum, a maximum and neither respectively. Return to a first-derivative sign chart or compare function values.
Optional enrichment, not required for the core method here: For sufficiently smooth functions, a higher-order test can help. If the first nonzero derivative at c after f itself has order n, with all earlier positive-order derivatives zero, an even n gives a minimum for a positive nth derivative and a maximum for a negative nth derivative. Odd n gives neither. This comes from the leading nonzero Taylor term; do not apply it without the required smoothness or when no such derivative is found.
A horizontal tangent is not necessarily an inflection point either. The graph must change concavity to have an inflection point under the usual smooth-graph definition. The fourth-power example stays concave upward around zero even though its second derivative vanishes there.
ORIGINAL WORKED EXAMPLE
Worked example 3 · Main-style trap
Classify every stationary point of f(x) = x⁴ − 4x³.
f′(x) = 4x²(x − 3), so the stationary points are 0 and 3.
f″(x) = 12x(x − 2). At x = 3 it is 36 > 0, giving a local minimum. At x = 0 it is zero, so no conclusion follows from this test.
Inspect f′: away from zero, x² is positive. Thus f′ is negative both just left and just right of zero; the graph continues decreasing. Zero is not an extremum.
At 3 the derivative changes from negative to positive. f(3) = 81 − 108 = −27. Because the quartic tends to +∞ at both ends and has no other turn, this is also the absolute minimum on the real line.
Answer x = 0 is stationary but not an extremum; minimum −27 at x = 3; no absolute maximum.
06 / UNDERSTAND & APPLY
6. Corners, holes and restricted intervals
Absolute values deserve a domain split before differentiation. Their algebraic rule changes when the expression inside the bars changes sign. A corner can be the best point even though no tangent slope exists there. Compare one-sided behaviour or use the definition directly.
Rational functions require exclusion of denominator zeros before searching for extrema. Multiplying an equation by a denominator can create an invalid candidate at a hole. Similarly, squaring an equation can introduce extraneous roots. Return every proposed solution to both the original equation and the domain.
On an open or unbounded domain, examine boundary limits and behaviour at infinity as well as finite critical points. A local minimum may coexist with values dropping without bound elsewhere. If the domain has disconnected pieces, analyse each piece and then compare across all admissible inputs.
ORIGINAL WORKED EXAMPLE
Worked example 4 · Advanced-style domain check
Find the absolute extrema of f(x) = |x − 1| + x²/4 on [−2,3].
For x < 1, f = 1 − x + x²/4, so f′ = −1 + x/2. Its putative zero x = 2 is outside this branch and must be rejected.
For x > 1, f = x − 1 + x²/4, so f′ = 1 + x/2. Its putative zero x = −2 is outside this branch and must be rejected.
The derivative is negative throughout [−2,1) and positive throughout (1,3]. Thus the corner x = 1 is a minimum.
Compare the included endpoints and corner: f(−2) = 4, f(1) = 1/4, f(3) = 17/4.
Answer Absolute minimum 1/4 at x = 1; maximum 17/4 at x = 3.
07 / UNDERSTAND & APPLY
7. Turn a story into one function
An optimization story usually starts with several variables. Choose a clear input, express the target quantity, then use the constraint to eliminate the other variables. The resulting domain comes from the original physical conditions. Differentiate only after that reduction.
A good solution contains four different justifications: why the formula models the problem, why the domain describes valid choices, why the candidate is mathematically optimal, and what the answer means with units. A correct derivative cannot rescue an incorrect perimeter equation.
For smooth positive targets, sometimes a simpler equivalent objective is available. Maximizing a positive quantity also maximizes its logarithm because logarithm is strictly increasing. Minimizing a nonnegative distance is equivalent to minimizing its square. State why the transformation preserves the ordering and keep the original domain.
Boundary checks can be physical rather than computational. A rectangle collapsing to zero width has zero area. If the target is positive at an interior critical point and rises then falls across the whole domain, the interior point is the global maximum. Do not merely say “f″ is negative, therefore absolute maximum” without such a global argument.
ORIGINAL WORKED EXAMPLE
Worked example 5 · Main-style modelling
A 48 m fence encloses three sides of a rectangle; a straight wall forms the fourth side. Find dimensions that maximize area, assuming the wall is long enough.
Let x be each side perpendicular to the wall and y the fenced side parallel to it. The constraint is 2x + y = 48, not 2x + 2y = 48.
Area A(x) = xy = x(48 − 2x). Nondegenerate rectangles require 0 < x < 24.
A′(x) = 48 − 4x vanishes at x = 12. It is positive before 12 and negative after 12, so this is the absolute maximum on the admissible interval.
Then y = 48 − 24 = 24 m. Area is 12 × 24 = 288 m². At the excluded degenerate boundaries, the area tends to zero, consistent with the result.
Answer Perpendicular sides 12 m each; parallel fenced side 24 m; maximum area 288 m².
08 / UNDERSTAND & APPLY
8. Original practice: decide before differentiating
These original questions are written for practice rather than taken from official papers. Main-style indicates a relatively direct application; Advanced-style indicates an additional domain, parameter or modelling decision. Try a complete solution before opening the explanation, and identify the exact point where your reasoning differs if your answer is wrong.
Suggested timed attempt, not an official exam time: allow 12 minutes for questions 1–4 and 24 minutes for questions 5–8. Work without the solutions, and mark any answer whose reasoning you cannot explain. After review, retry missed questions the next day without looking at your old working. Record whether the cause was a concept, domain or constraint choice, algebra, or units; speed comes after a correct method.
ORIGINAL PRACTICE · 1
1 · Foundation. Find absolute extrema of f(x) = x² − 6x + 5 on [1,5].
Show a hint
Include both endpoints as well as the vertex.
Reveal answer & reasoning
Minimum −4 at x = 3; maximum 0 at x = 1 and x = 5.
f′ = 2x − 6 gives x = 3. Candidate values are f(1) = 0, f(3) = −4 and f(5) = 0. Equal endpoint values can both attain the maximum.
ORIGINAL PRACTICE · 2
2 · Main-style. Does f(x) = x³ + 2 have a local extremum at zero? Does it have absolute extrema on the real line?
Show a hint
Inspect the derivative on both sides, then the tails.
Reveal answer & reasoning
No local extremum at zero; no absolute maximum or minimum.
f′ = 3x² is positive on both sides of zero and zero only at zero. The graph increases through its horizontal tangent. It tends to +∞ to the right and −∞ to the left.
ORIGINAL PRACTICE · 3
3 · Main-style. Find absolute extrema of f(x) = |2x + 1| on [−2,1].
Show a hint
Where does the absolute-value expression change rule?
Reveal answer & reasoning
Minimum 0 at −1/2; maximum 3 at −2 and 1.
The corner is at 2x + 1 = 0, or x = −1/2. The value there is zero, the smallest possible absolute value. Both endpoints give 3. The two linear pieces have no other stationary points.
ORIGINAL PRACTICE · 4
4 · Main-style. Find the greatest value of f(x) = x(6 − x) for 0 < x < 6. Does a least value exist?
Show a hint
An open interval need not prevent an interior maximum.
Reveal answer & reasoning
Greatest value 9 at x = 3; no least value.
f = 9 − (x − 3)², so it reaches 9 at 3. All allowed products are positive, yet approach zero near either excluded endpoint. Thus zero is an infimum, not a minimum.
ORIGINAL PRACTICE · 5
5 · Advanced-style. For f(x) = x³ − 3ax on the real line, determine the local extrema for every real parameter a.
Show a hint
Separate a > 0, a = 0 and a < 0 before solving.
Reveal answer & reasoning
For a > 0: local maximum 2a√a at −√a and local minimum −2a√a at √a. For a ≤ 0: no local extrema.
f′ = 3(x² − a). If a > 0, the derivative signs are +, −, + across −√a and √a; substitute these inputs into f. If a = 0 the function is x³ and has no turn at zero. If a < 0, x² − a is strictly positive everywhere.
ORIGINAL PRACTICE · 6
6 · Advanced-style. Find the minimum of f(x) = x + 9/x on x > 0, and state whether the same answer is a global minimum on x ≠ 0.
Show a hint
Do not carry an answer from one domain into another.
Reveal answer & reasoning
On x > 0 the minimum is 6 at x = 3. On x ≠ 0 there is no global minimum.
f′ = 1 − 9/x² is negative for 0 < x < 3 and positive for x > 3. Thus 3 is the positive-domain minimum, with value 6. On the enlarged domain, f tends to −∞ as x approaches zero from below, so no least value exists.
ORIGINAL PRACTICE · 7
7 · Advanced-style. A closed rectangular box has square base side x and volume 32 cm³. Find the dimensions giving the least total surface area.
Show a hint
Eliminate height using volume; x must be positive.
Reveal answer & reasoning
Base 32^(1/3) cm by 32^(1/3) cm; height 32^(1/3) cm.
Height h = 32/x². Total area S = 2x² + 4xh = 2x² + 128/x for x > 0. S′ = 4x − 128/x² = 4(x³ − 32)/x². It changes from negative to positive at x = 32^(1/3), and S tends to +∞ at both domain boundaries. Then h = 32/x² = x, so the optimal box is a cube.
ORIGINAL PRACTICE · 8
8 · Advanced-style. Find the largest value of f(x) = x²(1 − x)³ on [0,1].
Show a hint
Factor the derivative instead of expanding the polynomial.
Reveal answer & reasoning
Maximum 108/3125 at x = 2/5.
f′ = 2x(1 − x)³ − 3x²(1 − x)² = x(1 − x)²(2 − 5x). On the interior, its sign changes + to − at 2/5. Both endpoints give zero. Substitution gives (4/25)(27/125) = 108/3125. The endpoints are also absolute minima, with value zero.
09 / UNDERSTAND & APPLY
9. A compact revision method
When a solution fails, locate the type of error. A missing endpoint is a candidate-list error. A wrong interval for an absolute-value branch is a domain error. Calling every zero of f′ a maximum is a classification error. A correct x but wrong requested area is an interpretation error. Fixing the category is more useful than simply repeating the arithmetic.
For a fast final check, draw a rough graph from your derivative signs and mark every excluded input with an open point or a vertical break. The drawing need not be artistic. It should agree with your maximum/minimum labels, limiting behaviour and endpoint values.
To test understanding, explain why x⁴ and x³ behave differently at zero without saying “because I memorized it.” Their derivative signs reveal whether the graph reverses direction. Then explain why an absolute maximum can occur where the derivative is nonzero: an endpoint has only one admissible direction of travel.
Mastery develops by combining modelling, calculus and verification on unfamiliar problems. These notes provide a complete first pass and a diagnostic practice set; they are not a guarantee of exam performance. Revisit the particular section that explains any decision you could not justify.
Domain → candidate list → sign test or value comparison → interpreted answer.
At f″ = 0, stop the second-derivative test and change methods.
Local is a nearby comparison; absolute includes every admissible input.
If a value is only approached, check attainment before naming an extremum.
State both the extreme value and every input where it occurs.
KEEP THE BIG IDEAS
Quick revision
A stationary point is a candidate, not a verdict.
Corners and included endpoints can determine the answer.
The extreme value theorem needs continuity on a closed, bounded interval.
Optimization begins with a model and domain, and ends with a justified interpretation.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.