MathsISC Class 12 · Applications of Derivatives

Maxima and Minima: Critical Points, Derivative Tests and Absolute Extremes

Q: What is the difference between a local and an absolute maximum?

A local maximum is the highest value in a small neighbourhood of a point. An absolute maximum is the highest value over the whole domain or interval. A function can have several local maxima, and its absolute maximum may be at an end point rather than at a turning point.

Q: Which test should I use: first or second derivative?

The second derivative test is usually quicker. Use the first derivative test when f″(c) = 0, when f″ is hard to find, or when f′ does not exist at the point.

Q: Is every stationary point a maximum or minimum?

No. If f′ does not change sign, as for x³ at 0, the stationary point is a point of inflexion, not an extreme.

Q: Do I need the derivative test for absolute extremes on a closed interval?

No. Find the critical points inside the interval, evaluate f there and at both end points, and compare. The largest and smallest values are the absolute maximum and minimum.

Q: What does the ISC 2027 syllabus say about the tests?

It lists critical, stationary and turning points, absolute and local maxima and minima, and asks for the first derivative test motivated geometrically and the second derivative test given as a provable tool, with application problems.

Updated 15 October 2026 · Fact-checked

Local maxima and minima occur only at critical points, where f′(x) = 0 or f′ does not exist. A point is a local maximum if f′ changes from positive to negative there (or f″ < 0), and a local minimum if f′ changes from negative to positive (or f″ > 0). On a closed interval, compare the end points too.

At each turning point the tangent is flat and f′ changes sign: plus to minus gives a maximum, minus to plus gives a minimum.

Illustration drawn for QuestionBuddy.

Understand Maxima and Minima

Local and absolute extremes. $f$ has a local maximum at $c$ if $f (c) \geq f (x)$ for all $x$ near $c$ , and a local minimum if $f (c) \leq f (x)$ near $c$ . The absolute (global) maximum or minimum is the largest or smallest value over the whole domain or interval. A local maximum need not be the absolute maximum: a cubic like $x^{3} - 3 x$ has a local maximum at $x = - 1$ but takes much larger values for large $x$ .

Critical and stationary points. Points where $f^{'} (c) = 0$ are stationary points: the tangent is horizontal. Together with points where $f^{'}$ does not exist (corners), they are the critical points, the only places where a local extreme can occur in the interior of an interval. A stationary point where $f^{'}$ changes sign is a turning point. Not every stationary point is an extreme: $f (x) = x^{3}$ has $f^{'} (0) = 0$ , but $f$ keeps increasing through 0, so it has neither a maximum nor a minimum there (it is a point of inflexion).

Increasing and decreasing. $f$ is strictly increasing on an interval where $f^{'} (x) > 0$ , and strictly decreasing where $f^{'} (x) < 0$ . The critical points split the number line into intervals; the sign of $f^{'}$ on each tells you how the graph moves. The ISC 2025 examiners noted confusion between "increasing" and "strictly increasing" and with the use of brackets; a function can be increasing (non-decreasing) on a closed interval that includes a point where $f^{'} = 0$ .

First derivative test (motivated geometrically, as the syllabus says). At a critical point $c$ :

$f^{'}$ changes from $+$ to $-$ : the graph rises then falls, so local maximum.
$f^{'}$ changes from $-$ to $+$ : local minimum.
$f^{'}$ keeps the same sign: neither.

This test always works, even when $f^{''} (c) = 0$ or $f^{'}$ does not exist at $c$ .

Second derivative test (a provable tool). If $f^{'} (c) = 0$ :

$f^{''} (c) < 0$ : local maximum (the curve is bending downwards).
$f^{''} (c) > 0$ : local minimum.
$f^{''} (c) = 0$ : the test fails; go back to the first derivative test. For $x^{4}$ at 0 the test fails, but $f^{'}$ changes from $-$ to $+$ , so it is a minimum.

**Absolute extremes on a closed interval $[a, b]$ .** A continuous function on a closed interval always has an absolute maximum and minimum. To find them, list the values of $f$ at every critical point inside $(a, b)$ and at the end points $a$ and $b$ ; the largest is the absolute maximum and the smallest the absolute minimum. No derivative test is needed for this comparison.

Reading turning points from a graph. In graph-based questions (such as ISC 2025 Q9, on $f (x) = - x^{3} + 27 x - 2$ ), the turning points are the peaks and troughs. You find their coordinates by solving $f^{'} (x) = 0$ ; the slope is positive on rising parts and negative on falling parts. The examiners noted candidates who did not set the derivative equal to zero to find the turning points.

Maxima and Minima: formulas to remember

Stationary point: $f^{'} (c) = 0$; Critical points also include points where f′ does not exist.
Increasing / decreasing: $f^{'} > 0 \Rightarrow strictly increasing, f^{'} < 0 \Rightarrow strictly decreasing$; Check the sign of f′ on each interval between critical points.
First derivative test: $f^{'} : + \to - \Rightarrow max; - \to + \Rightarrow min$; No change of sign: neither.
Second derivative test: $f^{'} (c) = 0, f^{''} (c) < 0 \Rightarrow max; f^{''} (c) > 0 \Rightarrow min$; Inconclusive if f″(c) = 0.
Absolute extremes on [a, b]: $compare f (a), f (b), f (c_{1}), f (c_{2}), \dots$; c₁, c₂, … are the critical points inside the interval.
Cubic turning points: $f (x) = a x^{3} + b x^{2} + c x + d \Rightarrow f^{'} (x) = 3 a x^{2} + 2 b x + c$; Two turning points if the discriminant 4b² − 12ac > 0.

How to solve Maxima and Minima questions

Use this sequence for "find the local maxima and minima" and "find the absolute maximum and minimum" questions.

1Find $f^{'} (x)$ and simplify it into factors.
2Solve $f^{'} (x) = 0$ (and note any points where $f^{'}$ does not exist) to get the critical points.
3For local extremes, find $f^{''} (x)$ and evaluate it at each critical point; state max or min from its sign.
4If $f^{''} (c) = 0$ , make a sign table for $f^{'}$ around $c$ and use the first derivative test.
5Find the local maximum or minimum value by substituting $c$ into $f$ , not into $f^{'}$ .
6For a closed interval, evaluate $f$ at the end points as well, and pick the largest and smallest of all the values.
7For increasing/decreasing questions, use the sign of $f^{'}$ on each interval between critical points and write the intervals with correct brackets.

Maxima and Minima in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Absolute maximum on a closed interval

Suppose the temperature in a town on a spring day is modelled by $T (t) = 18 + 2 t - \frac{t ^{2}}{12}$ °C, where $t$ is the number of hours after midnight, $0 \leq t \leq 24$ .

Find the highest and lowest temperatures during the day and when they occur.

Try it first, then show the solutionSolution

$T^{'} (t) = 2 - \frac{t}{6} = 0 \Rightarrow t = 12$ .
$T^{''} (t) = - \frac{1}{6} < 0$ , so $t = 12$ is a maximum: $T (12) = 18 + 24 - 12 = 30$ °C.
End points: $T (0) = 18$ °C and $T (24) = 18 + 48 - 48 = 18$ °C.

Answer: The highest temperature is 30 °C at noon; the lowest is 18 °C, at midnight at both ends of the day.

Memory hook: On a closed interval, check the ends too: the coldest moments were at midnight.

Local maximum by the second derivative test

Imagine a stone thrown straight up from ground level, whose height after $t$ seconds is $h (t) = 20 t - 5 t^{2}$ metres.

Find the greatest height and show that it is a maximum.

Try it first, then show the solutionSolution

$h^{'} (t) = 20 - 10 t = 0 \Rightarrow t = 2$ s.
$h^{''} (t) = - 10 < 0$ : maximum.
$h (2) = 40 - 20 = 20$ m.

Answer: The stone reaches 20 m after 2 seconds.

Memory hook: At the top of the throw, velocity (h′) is zero and acceleration (h″) is negative.

Tricks to remember Maxima and Minima

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hill and valley
Up then down (+ to −) is a hill: maximum. Down then up (− to +) is a valley: minimum.
Use it when: The first derivative test.
Frown and smile
f″ < 0 looks like a frown ∩ (max); f″ > 0 like a smile ∪ (min).
Use it when: The second derivative test.
Ends are candidates
For absolute extremes on [a, b], the ends always join the competition.
Use it when: Closed-interval questions.
Zero slope is not enough
x³ has zero slope at 0 but no hill or valley. Check the sign change.
Use it when: Assertion-reason items.

Maxima and Minima: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Critical points of a cubic · 1 mark · MCQBoard level

The critical points of $f (x) = x^{3} - 3 x$ are:
(a) $x = 0$ only
(b) $x = \pm 1$
(c) $x = \pm 3$
(d) $x = 3$

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ .

Answer: (b) $x = \pm 1$ .

Example 2: A local minimum for x > 0 · 2 marks · Short answerBoard level

Show that $f (x) = x + \frac{1}{x}$ , $x > 0$ , has a local minimum at $x = 1$ , and find the minimum value.

Try it first, then show the solutionSolution

$f^{'} (x) = 1 - \frac{1}{x ^{2}} = 0 \Rightarrow x = 1$ (as $x > 0$ ). (1 mark)
$f^{''} (x) = \frac{2}{x ^{3}}$ , so $f^{''} (1) = 2 > 0$ : local minimum, with value $f (1) = 2$ . (1 mark)

Answer: Local minimum at $x = 1$ ; minimum value 2. (ISC 2025 asked a similar 2-mark proof for $lo g (1 + x) + \frac{1}{1 + x}$ ; the examiners noted candidates who failed to complete the test.)

Example 3: Local maximum and minimum of a cubic · 3 marks · Short answerHarder board level

Find the local maxima and minima of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ and the corresponding values.

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3) = 0 \Rightarrow x = 1, 3$ . (1 mark)
$f^{''} (x) = 6 x - 12$ : $f^{''} (1) = - 6 < 0$ (max), $f^{''} (3) = 6 > 0$ (min). (1 mark)
$f (1) = 1 - 6 + 9 + 1 = 5$ ; $f (3) = 27 - 54 + 27 + 1 = 1$ . (1 mark)

Answer: Local maximum 5 at $x = 1$ ; local minimum 1 at $x = 3$ .

Example 4: Absolute extremes on a closed interval · 3 marks · Short answerHarder board level

Find the absolute maximum and minimum values of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 1$ on $[0, 3]$ .

Try it first, then show the solutionSolution

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ ; critical points $x = 1, 2$ , both inside $[0, 3]$ . (1 mark)
$f (0) = 1$ , $f (1) = 6$ , $f (2) = 5$ , $f (3) = 54 - 81 + 36 + 1 = 10$ . (1½ marks)
Largest value 10, smallest value 1. (½ mark)

Answer: Absolute maximum 10 at $x = 3$ ; absolute minimum 1 at $x = 0$ . The local maximum $f (1) = 6$ is not the absolute maximum.

Example 5: When the second derivative test fails · 3 marks · Short answerHarder board level

Examine $f (x) = x^{4}$ and $g (x) = (x - 1)^{3}$ for local extremes at their stationary points.

Try it first, then show the solutionSolution

$f^{'} (x) = 4 x^{3} = 0 \Rightarrow x = 0$ ; $f^{''} (0) = 0$ , so the second derivative test fails. (1 mark)
First derivative test: $f^{'} < 0$ for $x < 0$ and $f^{'} > 0$ for $x > 0$ , so $f$ has a local minimum 0 at $x = 0$ . (1 mark)
$g^{'} (x) = 3 (x - 1)^{2} \geq 0$ on both sides of $x = 1$ : no sign change, so $g$ has neither a maximum nor a minimum there (a point of inflexion). (1 mark)

Answer: $x^{4}$ : local minimum at 0. $(x - 1)^{3}$ : no extreme at 1.

Example 6: Turning points from a graph · 5 marks · Case-basedHOTS / competency

The graph of $f (x) = x^{3} - 12 x + 5$ has a peak at A and a trough at B.
(i) Find the coordinates of A and B. (2 marks)
(ii) Write the intervals on which $f$ is strictly increasing and strictly decreasing. (1 mark)
(iii) Arrange $f^{''} (1)$ , $f (0)$ and $f^{'} (2)$ in ascending order. (1 mark)
(iv) Is the local maximum the absolute maximum of $f$ on $R$ ? (1 mark)

Try it first, then show the solutionSolution

(i) $f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x = \pm 2$ . $f^{''} (x) = 6 x$ : $f^{''} (- 2) < 0$ , so A $= (- 2, f (- 2)) = (- 2, 21)$ ; $f^{''} (2) > 0$ , so B $= (2, - 11)$ .
(ii) $f^{'} > 0$ on $(- \infty, - 2) \cup (2, \infty)$ (strictly increasing); $f^{'} < 0$ on $(- 2, 2)$ (strictly decreasing).
(iii) $f^{''} (1) = 6$ , $f (0) = 5$ , $f^{'} (2) = 0$ . Ascending: $f^{'} (2) < f (0) < f^{''} (1)$ .
(iv) No: $f (x) \to \infty$ as $x \to \infty$ (for example $f (5) = 70 > 21$ ), so there is no absolute maximum on $R$ .

Answer: A $(- 2, 21)$ , B $(2, - 11)$ ; increasing on $(- \infty, - 2) \cup (2, \infty)$ , decreasing on $(- 2, 2)$ ; $0 < 5 < 6$ ; the local maximum is not absolute.

Where students lose marks in Maxima and Minima

The slips examiners see most often, and what to write instead.

Mistake: Not completing the derivative test
The ISC 2025 examiners noted candidates who failed to prove that f(x) attains its minimum at x = 0, that is, who did not complete a derivative test.
Do this: After solving f′(x) = 0, always state f″(c) with its sign, or a sign table for f′, and conclude in words.
Usually costs: 1 mark in a 2-mark proof.
Mistake: Substituting into f′ instead of f for the value
The extreme value is f(c); f′(c) is always 0.
Do this: Write "maximum value = f(c) =" before substituting.
Usually costs: 1 mark.
Mistake: Ignoring end points
On a closed interval the absolute extreme may be at an end, as in f(x) = 2x³ − 9x² + 12x + 1 on [0, 3].
Do this: List f at every critical point and at both ends, then pick the largest and smallest.
Usually costs: 1 to 2 marks.
Mistake: Confusing increasing with strictly increasing
The 2025 examiners reported wrong brackets and confusion between the two terms.
Do this: Strictly increasing uses f′ > 0 on an open interval; increasing allows f′ = 0 at isolated points and may use closed brackets.
Usually costs: 1 mark in Section A.

Exam tips

Factorise $f^{'} (x)$ completely before solving; it makes the critical points and the sign table obvious.
Write "since $f^{''} (c) < 0$ , $f$ has a local maximum at $x = c$ " in full; examiners look for the reason.
For graph-based questions, label the turning points with both coordinates.
If asked to arrange values in order, compute each one separately first (for example $f^{''} (1)$ , $f (0)$ , $f^{'} (2)$ ), then order them.
Remember that a local maximum value can be smaller than a local minimum value of the same function.

Practice questions on Maxima and Minima

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The function $f (x) = x^{3} - 27 x$ is strictly increasing on:
(a) $(- 3, 3)$
(b) $(- \infty, - 3) \cup (3, \infty)$
(c) $(- 3, \infty)$
(d) $(- \infty, 3)$
Hint
$f^{'} (x) = 3 x^{2} - 27$ .
Answer
(b).
Q2 · 2 marks · Short answerBoard level
Find the maximum value of $f (x) = sin x + cos x$ for $0 < x < \frac{π}{2}$ .
Hint
Solve $cos x - sin x = 0$ .
Answer
$f^{'} (x) = cos x - sin x = 0 \Rightarrow x = \frac{π}{4}$ ; $f^{''} (\frac{π}{4}) = - 2 < 0$ . Maximum $2$ .
Q3 · 3 marks · Short answerHarder board level
Find the local maximum value of $f (x) = x e^{- x}$ .
Hint
Product rule.
Answer
$f^{'} (x) = e^{- x} (1 - x) = 0 \Rightarrow x = 1$ . $f^{''} (x) = e^{- x} (x - 2)$ , $f^{''} (1) = - \frac{1}{e} < 0$ . Local maximum $f (1) = \frac{1}{e}$ .
Q4 · 3 marks · Short answerHarder board level
Find the absolute maximum and minimum of $f (x) = 4 x - \frac{x ^{2}}{2}$ on $[- 2, \frac{9}{2}]$ .
Hint
One critical point; then compare with the ends.
Answer
$f^{'} (x) = 4 - x = 0 \Rightarrow x = 4$ . $f (- 2) = - 10$ , $f (4) = 8$ , $f (4.5) = 18 - 10.125 = 7.875$ . Absolute maximum 8, absolute minimum $- 10$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $f (x) = x^{3}$ has no local extremum at $x = 0$ .
Reason (R): $f^{'} (0) = 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Does $f^{'}$ change sign at 0?
Answer
(b). Both are true, but a zero derivative does not explain the absence of an extreme; the reason is that $f^{'}$ does not change sign.
Q6 · 3 marks · Short answerHOTS / competency
Find $a$ and $b$ if $f (x) = x^{3} + a x^{2} + b x$ has turning points at $x = - 1$ and $x = 3$ , and say which is a maximum.
Hint
$- 1$ and $3$ are the roots of $f^{'} (x) = 0$ .
Answer
$f^{'} (x) = 3 x^{2} + 2 a x + b$ . Sum of roots $2 = - \frac{2 a}{3} \Rightarrow a = - 3$ ; product $- 3 = \frac{b}{3} \Rightarrow b = - 9$ . $f^{''} (x) = 6 x - 6$ : $f^{''} (- 1) = - 12 < 0$ (maximum, value 5); $f^{''} (3) = 12 > 0$ (minimum, value $- 27$ ).
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a section of a roller-coaster track follows $h (x) = x^{3} - 9 x^{2} + 24 x + 2$ metres above the ground, for $0 \leq x \leq 6$ , where $x$ is the horizontal distance in tens of metres.
(i) Find the turning points of the track. (1 mark)
(ii) Classify them using the second derivative. (1 mark)
(iii) Find the highest and lowest points of this section. (1 mark)
(iv) On which interval does the track go downhill? (1 mark)
Hint
$h^{'} (x) = 3 (x - 2) (x - 4)$ .
Answer
(i) $x = 2$ and $x = 4$ : $(2, 22)$ and $(4, 18)$ . (ii) $h^{''} (x) = 6 x - 18$ : $h^{''} (2) = - 6$ (local max), $h^{''} (4) = 6$ (local min). (iii) $h (0) = 2$ , $h (6) = 38$ ; highest point 38 m at $x = 6$ , lowest 2 m at $x = 0$ . (iv) Downhill where $h^{'} < 0$ : $2 < x < 4$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a critical point?Show answer
A point where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist.
2.State the first derivative test for a minimum.Show answer
If $f^{'}$ changes from negative to positive at $c$ , then $f$ has a local minimum at $c$ .
3.When does the second derivative test fail?Show answer
When $f^{''} (c) = 0$ .
4.Find the stationary point of $f (x) = x^{2} - 8 x$ .Show answer
$x = 4$ (a minimum, value $- 16$ ).
5.How do you find the absolute maximum on $[a, b]$ ?Show answer
Compare $f$ at the critical points inside and at $a$ and $b$ ; take the largest.
6.Does $f (x) = x^{3}$ have a local extreme at 0?Show answer
No; $f^{'}$ does not change sign there.

Maxima and Minima: frequently asked questions

What is the difference between a local and an absolute maximum?

A local maximum is the highest value in a small neighbourhood of a point. An absolute maximum is the highest value over the whole domain or interval. A function can have several local maxima, and its absolute maximum may be at an end point rather than at a turning point.

Which test should I use: first or second derivative?

The second derivative test is usually quicker. Use the first derivative test when f″(c) = 0, when f″ is hard to find, or when f′ does not exist at the point.

Is every stationary point a maximum or minimum?

No. If f′ does not change sign, as for x³ at 0, the stationary point is a point of inflexion, not an extreme.

Do I need the derivative test for absolute extremes on a closed interval?

No. Find the critical points inside the interval, evaluate f there and at both end points, and compare. The largest and smallest values are the absolute maximum and minimum.

What does the ISC 2027 syllabus say about the tests?

It lists critical, stationary and turning points, absolute and local maxima and minima, and asks for the first derivative test motivated geometrically and the second derivative test given as a provable tool, with application problems.