MathsCBSE Class 12 · Application of Derivatives

Increasing and Decreasing Functions: Intervals by the Derivative Test (Class 12)

Updated 15 October 2026 · Fact-checked

A differentiable function is strictly increasing on an interval where f′(x) > 0 and strictly decreasing where f′(x) < 0. To find the intervals, solve f′(x) = 0, mark these critical points on a number line, test the sign of f′ in each part, and write the intervals. A constant function has f′(x) = 0 throughout.

Where f′ is positive the graph rises, where f′ is negative it falls; the critical points x = −1 and x = 1 split the number line into test intervals.

Illustration drawn for QuestionBuddy.

Understand Increasing and Decreasing Functions

A function $f$ is increasing on an interval if $x_{1} < x_{2}$ implies $f (x_{1}) \leq f (x_{2})$ , and strictly increasing if it implies $f (x_{1}) < f (x_{2})$ . Decreasing and strictly decreasing are defined the same way with the inequalities reversed. Graphically: an increasing graph rises from left to right.

The derivative test (NCERT Theorem in Section 6.3). Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$ .

If $f^{'} (x) > 0$ for every $x \in (a, b)$ , then $f$ is strictly increasing on $[a, b]$ .
If $f^{'} (x) < 0$ for every $x \in (a, b)$ , then $f$ is strictly decreasing on $[a, b]$ .
If $f^{'} (x) = 0$ for every $x \in (a, b)$ , then $f$ is constant on $[a, b]$ .

The reason is that $f^{'} (x)$ is the slope: positive slope everywhere means the graph keeps rising. Note that the converse is not exact: $f (x) = x^{3}$ is strictly increasing on $R$ although $f^{'} (0) = 0$ . An isolated zero of $f^{'}$ does not stop a function from being strictly increasing.

Finding the intervals.

1. Find $f^{'} (x)$ and factorise it.
2. Solve $f^{'} (x) = 0$ (and note points where $f^{'}$ is undefined, but inside the domain). These critical points split the domain into intervals.
3. On each interval, the sign of $f^{'}$ is constant; test one point or use the signs of the factors.
4. Positive sign: increasing; negative sign: decreasing. Write the answer in interval notation; closed brackets at critical points are acceptable for 'increasing' and 'decreasing'.

For trigonometric functions on an interval like $[0, 2 π]$ , first rewrite $f^{'}$ in a single-term form (for example $cos x + 3 sin x = 2 sin (x + \frac{π}{6})$ ) so that its zeros and signs are easy to read.

Proving monotonicity. To show a function is increasing on all of $R$ , show $f^{'} (x) > 0$ for all $x$ , often by writing $f^{'}$ as a sum of squares or exponentials plus a positive constant. For a quadratic $f^{'} (x) = p x^{2} + q x + r$ to be non-negative everywhere, you need $p > 0$ and discriminant $\leq 0$ .

Inequalities by monotonicity. To prove $g (x) > h (x)$ for $x > 0$ , set $F (x) = g (x) - h (x)$ . If $F (0) = 0$ and $F^{'} (x) > 0$ for $x > 0$ , then $F$ is increasing, so $F (x) > F (0) = 0$ . This gives, for example, $\frac{x}{1 + x} < lo g (1 + x) < x$ for $x > 0$ .

Monotonicity questions appear in nearly every board paper as 1-mark MCQs and 2- or 3-mark questions; the same sign chart also feeds the first derivative test for maxima and minima.

Increasing and Decreasing Functions: formulas to remember

Strictly increasing: $f^{'} (x) > 0 on (a, b) \Rightarrow f ↑ on [a, b]$; f continuous on $[a, b]$ , differentiable on $(a, b)$ .
Strictly decreasing: $f^{'} (x) < 0 on (a, b) \Rightarrow f ↓ on [a, b]$; Same conditions.
Constant: $f^{'} (x) = 0 on (a, b) \Rightarrow f constant on [a, b]$; Zero slope everywhere.
Critical points: $f^{'} (c) = 0 or f^{'} (c) undefined$; They split the domain into test intervals.
Quadratic derivative never negative: $p x^{2} + q x + r \geq 0 \forall x ⟺ p > 0, q^{2} - 4 p r \leq 0$; Used in 'increasing on R for which k' questions.
Inequality via monotonicity: $F (0) = 0, F^{'} (x) > 0 for x > 0 \Rightarrow F (x) > 0 for x > 0$; Set $F = g - h$ to prove $g > h$ .

How to solve Increasing and Decreasing Functions questions

Use this routine for 'find the intervals in which f is increasing or decreasing'.

1Write the domain of $f$ (all reals, $x > 0$ for logs, a given interval for trig functions).
2Differentiate and factorise $f^{'} (x)$ completely.
3Solve $f^{'} (x) = 0$ to get the critical points; arrange them in order on a number line.
4Choose a test value in each interval (or use the signs of factors) to find the sign of $f^{'}$ .
5Write: ' $f^{'} (x) > 0$ on ..., so f is strictly increasing on ...' and ' $f^{'} (x) < 0$ on ..., so f is strictly decreasing on ...'.
6For 'increasing on $R$ ' questions, show $f^{'} (x) > 0$ (or $\geq 0$ with isolated zeros) for every real $x$ .

Increasing and Decreasing Functions 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.

Increasing then decreasing

Suppose a cricket ball is hit straight up and its height after $t$ seconds is $h (t) = 20 t - 5 t^{2}$ metres, for $0 \leq t \leq 4$ .

On which time interval is the ball rising, and on which is it falling? What is the greatest height?

Try it first, then show the solutionSolution

$h^{'} (t) = 20 - 10 t = 10 (2 - t)$ .
$h^{'} (t) > 0$ for $0 < t < 2$ : the height increases on $[0, 2]$ .
$h^{'} (t) < 0$ for $2 < t < 4$ : the height decreases on $[2, 4]$ .
Greatest height at $t = 2$ : $h (2) = 40 - 20 = 20$ m.

Answer: Rising on $[0, 2]$ s, falling on $[2, 4]$ s; greatest height 20 m at $t = 2$ s.

Memory hook: Positive slope: going up; negative slope: coming down.

Always increasing, more and more slowly

Imagine a phone battery charging so that its charge (in %) after $t$ hours is $C (t) = 100 (1 - e^{- t /2})$ .

Show that the charge always increases. Find the charging rate at $t = 2$ and say what happens to the rate as time passes.

Try it first, then show the solutionSolution

$C^{'} (t) = 100 \cdot \frac{1}{2} e^{- t /2} = 50 e^{- t /2}$ .
$e^{- t /2} > 0$ for all $t$ , so $C^{'} (t) > 0$ : the charge is strictly increasing.
$C^{'} (2) = 50 e^{- 1} \approx 18.4$ % per hour.
$C^{'} (t) = 50 e^{- t /2}$ itself decreases as $t$ grows, so charging slows down near full charge.

Answer: C is strictly increasing since $C^{'} (t) = 50 e^{- t /2} > 0$ ; at $t = 2$ the rate is $\frac{50}{e} \approx 18.4$ % per hour, and it keeps falling.

Memory hook: Increasing does not mean speeding up: the battery still charges, just more slowly.

Tricks to remember Increasing and Decreasing Functions

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

Sign of the slope
$f^{'} > 0$ : uphill (increasing). $f^{'} < 0$ : downhill (decreasing). $f^{'} = 0$ : flat.
Use it when: Every monotonicity question.
Zeros split, signs decide
Critical points cut the number line; one test value per piece decides the sign of the whole piece.
Use it when: Building the sign chart.
x³ is the exception
A single zero of $f^{'}$ (like $x^{3}$ at 0) does not stop strict increase. Remember $x^{3}$ when an assertion says ' $f^{'} > 0$ everywhere'.
Use it when: Assertion-reason traps.
Compare with zero via F = g − h
To prove $g > h$ , show $g - h$ starts at 0 and increases.
Use it when: Inequality proofs.

Increasing and Decreasing Functions: solved examples

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

Example 1: Intervals for a cubic · 3 marks · Short answerBoard level

Find the intervals in which $f (x) = x^{3} - 6 x^{2} + 9 x + 4$ is strictly increasing or strictly decreasing.

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ . (1 mark)
$f^{'} (x) = 0$ at $x = 1, 3$ , splitting $R$ into $(- \infty, 1)$ , $(1, 3)$ , $(3, \infty)$ . (½ mark)
Signs: $x = 0$ gives $f^{'} = 9 > 0$ ; $x = 2$ gives $f^{'} = - 3 < 0$ ; $x = 4$ gives $f^{'} = 9 > 0$ . (1 mark)
Strictly increasing on $(- \infty, 1]$ and $[3, \infty)$ ; strictly decreasing on $[1, 3]$ . (½ mark)

Answer: Increasing on $(- \infty, 1] \cup [3, \infty)$ ; decreasing on $[1, 3]$ .

Example 2: Increasing on the whole real line · 2 marks · Very short answerBoard level

Show that $f (x) = e^{2 x} + x$ is strictly increasing on $R$ .

Try it first, then show the solutionSolution

$f^{'} (x) = 2 e^{2 x} + 1$ . (1 mark)
$e^{2 x} > 0$ for all real $x$ , so $f^{'} (x) > 1 > 0$ ; hence f is strictly increasing on $R$ . (1 mark)

Answer: $f^{'} (x) = 2 e^{2 x} + 1 > 0$ for all x.

Example 3: A trigonometric function on [0, 2π] · 3 marks · Short answerHarder board level

Find the intervals in $[0, 2 π]$ on which $f (x) = sin x - 3 cos x$ is strictly increasing or strictly decreasing.

Try it first, then show the solutionSolution

$f^{'} (x) = cos x + 3 sin x = 2 sin (x + \frac{π}{6})$ . (1 mark)
$f^{'} (x) = 0$ when $x + \frac{π}{6} = π$ or $2 π$ , i.e. $x = \frac{5 π}{6}$ or $\frac{11 π}{6}$ . (1 mark)
$f^{'} > 0$ when $x + \frac{π}{6} \in (0, π)$ or $(2 π, \frac{13 π}{6})$ , i.e. on $[0, \frac{5 π}{6})$ and $(\frac{11 π}{6}, 2 π]$ ; $f^{'} < 0$ on $(\frac{5 π}{6}, \frac{11 π}{6})$ . (1 mark)

Answer: Increasing on $[0, \frac{5 π}{6}]$ and $[\frac{11 π}{6}, 2 π]$ ; decreasing on $[\frac{5 π}{6}, \frac{11 π}{6}]$ .

Example 4: Finding a parameter · 3 marks · Short answerHarder board level

Find the values of $k$ for which $f (x) = k x^{3} - 9 x^{2} + 9 x + 3$ is increasing on $R$ .

Try it first, then show the solutionSolution

$f^{'} (x) = 3 k x^{2} - 18 x + 9 = 3 (k x^{2} - 6 x + 3)$ . (½ mark)
We need $k x^{2} - 6 x + 3 \geq 0$ for all $x$ . If $k \leq 0$ this fails for large positive $x$ (or at $x = 1$ when $k = 0$ : $- 3 < 0$ ). So $k > 0$ . (1 mark)
A quadratic with positive leading coefficient is never negative when its discriminant is $\leq 0$ : $36 - 12 k \leq 0$ , so $k \geq 3$ . (1 mark)
At $k = 3$ , $f^{'} (x) = 9 (x - 1)^{2} \geq 0$ with a single zero, so f is still increasing. (½ mark)

Answer: $k \geq 3$

Example 5: Proving an inequality by monotonicity · 3 marks · Short answerHOTS / competency

Prove that $\frac{x}{1 + x} < lo g (1 + x) < x$ for all $x > 0$ .

Try it first, then show the solutionSolution

Let $g (x) = x - lo g (1 + x)$ . Then $g^{'} (x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x} > 0$ for $x > 0$ , and $g (0) = 0$ . So $g$ is strictly increasing and $g (x) > 0$ : $lo g (1 + x) < x$ . (1½ marks)
Let $h (x) = lo g (1 + x) - \frac{x}{1 + x}$ . Then $h^{'} (x) = \frac{1}{1 + x} - \frac{1}{( 1 + x ) ^{2}} = \frac{x}{( 1 + x ) ^{2}} > 0$ for $x > 0$ , and $h (0) = 0$ . So $h (x) > 0$ : $lo g (1 + x) > \frac{x}{1 + x}$ . (1½ marks)

Answer: Both differences start at 0 and increase for $x > 0$ , giving $\frac{x}{1 + x} < lo g (1 + x) < x$ .

Example 6: Case study: a company's profit over output · 4 marks · Case-basedHOTS / competency

Suppose a small firm's weekly profit (in thousand rupees) when it makes $x$ hundred units is $P (x) = - x^{3} + 12 x^{2} - 36 x + 80$ , for $0 \leq x \leq 8$ .
(i) Find $P^{'} (x)$ and its zeros. (1 mark)
(ii) Find the intervals where profit increases and decreases. (2 marks)
(iii) Compare $P (2)$ , $P (6)$ and $P (8)$ . (1 mark)

Try it first, then show the solutionSolution

(i) $P^{'} (x) = - 3 x^{2} + 24 x - 36 = - 3 (x - 2) (x - 6)$ ; zeros at $x = 2$ and $x = 6$ . (1 mark)
(ii) For $0 < x < 2$ : $P^{'} < 0$ (decreasing). For $2 < x < 6$ : $P^{'} > 0$ (increasing). For $6 < x < 8$ : $P^{'} < 0$ (decreasing). (2 marks)
(iii) $P (2) = - 8 + 48 - 72 + 80 = 48$ , $P (6) = - 216 + 432 - 216 + 80 = 80$ , $P (8) = - 512 + 768 - 288 + 80 = 48$ . (1 mark)

Answer: (i) $x = 2, 6$ . (ii) Decreasing on $[0, 2]$ and $[6, 8]$ , increasing on $[2, 6]$ . (iii) $P (2) = 48$ , $P (6) = 80$ , $P (8) = 48$ thousand rupees.

Where students lose marks in Increasing and Decreasing Functions

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

Mistake: Not factorising f′(x)
Without factors, the sign in each interval is guesswork.
Wrong working
$f^{'} (x) = 3 x^{2} - 12 x + 9$ , so increasing when $x > 0$ .
Right working
$f^{'} (x) = 3 (x - 1) (x - 3)$ : positive outside $[1, 3]$ , negative inside.
Do this: Always factorise and use a sign chart.
Usually costs: The final intervals (1 to 2 marks).
Mistake: Testing only one interval
The sign of $f^{'}$ can change at every critical point.
Wrong working
$f^{'} (2) < 0$ , so f is decreasing everywhere.
Right working
Test each of $(- \infty, 1)$ , $(1, 3)$ , $(3, \infty)$ separately.
Do this: One test value per interval.
Usually costs: 1 mark.
Mistake: Ignoring the given domain
For trigonometric functions only the stated interval (such as $[0, 2 π]$ ) counts.
Wrong working
Increasing on $(- \frac{π}{6}, \frac{5 π}{6})$ .
Right working
Within $[0, 2 π]$ : increasing on $[0, \frac{5 π}{6}]$ and $[\frac{11 π}{6}, 2 π]$ .
Do this: Intersect your intervals with the given domain.
Usually costs: Half to one mark.
Mistake: Using f′ > 0 for a parameter question
Requiring strict $>$ for all x can wrongly exclude the boundary value.
Wrong working
$k x^{2} - 6 x + 3 > 0$ for all x gives $k > 3$ .
Right working
Increasing allows isolated zeros: $k \geq 3$ (at $k = 3$ , $f^{'} = 9 (x - 1)^{2} \geq 0$ ).
Do this: Check the boundary value separately.
Usually costs: The final answer mark.

Exam tips

1-mark MCQs on increasing or decreasing intervals appeared in the 2023, 2024 and 2025 board papers, and 2-mark interval questions in 2024, 2025 and 2026; practise fast sign charts.
Always write the factorised derivative and the critical points before the intervals.
Use interval notation; the board accepts closed brackets at critical points.
For trigonometric derivatives, convert $a cos x + b sin x$ to a single sine or cosine to read the sign easily.
In proofs on $R$ , give a reason such as ' $e^{x} > 0$ ' or 'a square is non-negative'.

Practice questions on Increasing and Decreasing Functions

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 $f (x) = x^{3} + 3 x$ is:
(a) increasing on $R$
(b) decreasing on $R$
(c) increasing only for $x > 0$
(d) neither increasing nor decreasing
Hint
$f^{'} (x) = 3 x^{2} + 3$ .
Answer
(a). $f^{'} (x) = 3 x^{2} + 3 > 0$ for all x.
Q2 · 2 marks · Very short answerBoard level
Find the intervals in which $f (x) = x^{2} - 4 x + 7$ is increasing or decreasing.
Hint
$f^{'} (x) = 2 (x - 2)$ .
Answer
$f^{'} (x) = 2 x - 4$ . $f^{'} < 0$ for $x < 2$ and $f^{'} > 0$ for $x > 2$ : decreasing on $(- \infty, 2]$ , increasing on $[2, \infty)$ .
Q3 · 2 marks · Very short answerBoard level
Show that $f (x) = 3 x + cos x$ is strictly increasing on $R$ .
Hint
$sin x \leq 1$ .
Answer
$f^{'} (x) = 3 - sin x \geq 3 - 1 = 2 > 0$ for all x, so f is strictly increasing.
Q4 · 3 marks · Short answerHarder board level
Find the intervals in which $f (x) = x^{4} - 8 x^{2} + 2$ is strictly increasing or strictly decreasing.
Hint
$f^{'} (x) = 4 x (x - 2) (x + 2)$ .
Answer
Critical points $- 2, 0, 2$ . Signs of $f^{'}$ : $(- \infty, - 2)$ negative, $(- 2, 0)$ positive, $(0, 2)$ negative, $(2, \infty)$ positive. Increasing on $[- 2, 0]$ and $[2, \infty)$ ; decreasing on $(- \infty, - 2]$ and $[0, 2]$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $f (x) = x^{3}$ is strictly increasing on $R$ .
Reason (R): $f^{'} (x) > 0$ for all $x \in R$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find $f^{'} (0)$ .
Answer
(c). $x_{1} < x_{2} \Rightarrow x_{1}^{3} < x_{2}^{3}$ , so A is true; but $f^{'} (0) = 0$ , so R is false.
Q6 · 2 marks · Very short answerHarder board level
Find the intervals in which $f (x) = x e^{- x}$ is increasing or decreasing.
Hint
$f^{'} (x) = e^{- x} (1 - x)$ and $e^{- x} > 0$ .
Answer
$f^{'} (x) = e^{- x} (1 - x)$ : positive for $x < 1$ , negative for $x > 1$ . Increasing on $(- \infty, 1]$ , decreasing on $[1, \infty)$ .
Q7 · 3 marks · Short answerHOTS / competency
(i) Find the intervals of increase and decrease of $f (x) = \frac{l o g x}{x}$ , $x > 0$ . (ii) Hence decide which is larger, $e^{π}$ or $π^{e}$ .
Hint
Since $π > e$ and f decreases after e, $f (π) < f (e)$ .
Answer
(i) $f^{'} (x) = \frac{1 - l o g x}{x ^{2}}$ : positive for $0 < x < e$ , negative for $x > e$ . Increasing on $(0, e]$ , decreasing on $[e, \infty)$ . (ii) $π > e$ gives $\frac{l o g π}{π} < \frac{l o g e}{e} = \frac{1}{e}$ , so $e lo g π < π$ , i.e. $lo g π^{e} < lo g e^{π}$ . Hence $e^{π} > π^{e}$ .

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.If $f^{'} (x) < 0$ on $(a, b)$ , what can you say about f?Show answer
f is strictly decreasing on $[a, b]$ (given continuity on $[a, b]$ ).
2.What is a critical point?Show answer
A point in the domain where $f^{'} (x) = 0$ or $f^{'}$ does not exist.
3.Is $x^{3}$ strictly increasing even though $f^{'} (0) = 0$ ?Show answer
Yes.
4.Where is $f (x) = x^{2}$ decreasing?Show answer
On $(- \infty, 0]$ .
5.Is $e^{x}$ increasing or decreasing on $R$ ?Show answer
Strictly increasing, since $(e^{x})^{'} = e^{x} > 0$ .
6.Which is larger, $e^{π}$ or $π^{e}$ ?Show answer
$e^{π}$ .

Increasing and Decreasing Functions: frequently asked questions

How do I find intervals where a function is increasing or decreasing?

Differentiate, factorise f′(x), find where it is zero, and check its sign in each interval between these points. Where f′ is positive the function increases; where it is negative the function decreases.

Can a function be strictly increasing if its derivative is zero somewhere?

Yes, if the derivative is zero only at isolated points and positive elsewhere. For example, x³ is strictly increasing on all real numbers although its derivative is zero at x = 0.

Should I use open or closed intervals?

Either is accepted at the critical points for increasing and decreasing functions in board answers, as long as the intervals are correct. Closed intervals like [1, 3] are common in NCERT.

Which NCERT section covers this topic?

Section 6.3 (Increasing and Decreasing Functions) of Chapter 6, Application of Derivatives, in NCERT Mathematics Part I for Class 12.

How is this topic linked to maxima and minima?

Where a function changes from increasing to decreasing it has a local maximum, and where it changes from decreasing to increasing it has a local minimum. This is the first derivative test.