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Questions on increasing and decreasing functions are among the quicker marks in Class 12 calculus, yet they are often lost to small slips: a sign error in factorising, an interval written with a union it should not have, or a function called non-increasing because its derivative touches zero. This post gives a four-step method and works six examples that cover the common question types.
Where is this in the syllabus?
Section 6.3 of the NCERT Class 12 chapter on application of derivatives covers it. The CBSE Mathematics curriculum for 2026-27 lists "increasing/decreasing functions" under applications of derivatives. The JEE (Main) 2026 syllabus calls it "monotonic-Increasing and decreasing functions", and the JEE Advanced 2026 syllabus includes it alongside tangents, normals and the mean value theorem. Check the current year's syllabus.
What does NCERT's theorem say?
NCERT first defines increasing and decreasing on an interval through values: f is strictly increasing on I if x₁ < x₂ in I always gives f(x₁) < f(x₂). Then Theorem 1 links this to the derivative. If f is continuous on [a, b] and differentiable on (a, b):
f is increasing on [a, b] if f′(x) > 0 for each x in (a, b);
f is decreasing on [a, b] if f′(x) < 0 for each x in (a, b);
f is constant on [a, b] if f′(x) = 0 for each x in (a, b).
The proof uses the mean value theorem. The conditions matter: the theorem works on one interval at a time, which is the root of the union trap below.
The four-step method
Find f′(x) and factorise it fully.
Mark the points where f′(x) = 0 or is undefined, and any points outside the domain.
Test the sign of f′ on each interval between those points by picking one value.
Write the intervals of increase and decrease separately.
Six worked examples
1. A cubic. f(x) = x³ − 6x² + 9x + 1. f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Test x = 0: f′ = 9 > 0. Test x = 2: f′ = 12 − 24 + 9 = −3 < 0. Test x = 4: f′ = 48 − 48 + 9 = 9 > 0. Strictly increasing on (−∞, 1] and [3, ∞); strictly decreasing on [1, 3]. Sense check: f(1) = 5 and f(3) = 1, so the function does fall between 1 and 3.
2. A zero derivative that does not stop growth. f(x) = x³. Here f′(0) = 0, which tempts students to say f is not increasing at 0. Use the definition instead: if x₁ < x₂ then x₁³ < x₂³, so f is strictly increasing on all of R. A derivative that is zero only at isolated points does not break strict increase.
3. The union trap. f(x) = x + 1/x, x ≠ 0. f′(x) = 1 − 1/x² = (x² − 1)/x². The sign is set by x² − 1, since x² > 0. So f′ < 0 for −1 < x < 0 and for 0 < x < 1. Answer: decreasing on (−1, 0) and on (0, 1), separately. It is not decreasing on (−1, 0) ∪ (0, 1): f(−0.5) = −2.5 and f(0.5) = 2.5, so a larger x gives a larger value across the gap. It is increasing on (−∞, −1] and [1, ∞).
4. A trigonometric function. f(x) = sin x + cos x on [0, 2π]. f′(x) = cos x − sin x, zero at x = π/4 and 5π/4. Test x = 0: f′ = 1 > 0. Test x = π/2: f′ = −1 < 0. Test x = 3π/2: f′ = 0 − (−1) = 1 > 0. Increasing on [0, π/4] and [5π/4, 2π]; decreasing on [π/4, 5π/4].
5. Finding a parameter. For which a is f(x) = x³ + ax² + 3x increasing on R? f′(x) = 3x² + 2ax + 3. We need f′(x) ≥ 0 for every x, with zeros only at isolated points. A quadratic with positive leading coefficient stays non-negative when its discriminant is at most zero: (2a)² − 4 × 3 × 3 ≤ 0, so 4a² ≤ 36 and −3 ≤ a ≤ 3. Check the edge a = 3: f′(x) = 3x² + 6x + 3 = 3(x + 1)², zero only at x = −1, so f is still strictly increasing, as in example 2.
6. Proving an inequality. Show that ln(1 + x) < x for x > 0. Let g(x) = x − ln(1 + x). Then g′(x) = 1 − 1/(1 + x) = x/(1 + x), which is positive for x > 0. So g is strictly increasing on [0, ∞), and g(0) = 0. Hence g(x) > 0 for x > 0, which is the inequality.
Which mistakes cost marks?
Factorising carelessly. One sign slip flips every interval. Test one value in each interval rather than trusting a pattern.
Joining intervals across a gap, as in example 3.
Forgetting the domain. For ln x or 1/x, mark excluded points before testing signs.
Treating f′ = 0 at a point as a change of direction. Check the sign on both sides, as in example 2.
Using f″. The second derivative is for concavity and the second derivative test, not for where f rises.
A practice routine
Do two polynomials, two rational functions and two trigonometric functions with the four-step method.
Do one parameter question and one inequality proof each week.
For every answer, check one value of f on each side of a turning point.
Then move on to maxima, minima and rates of change, which uses the same sign chart, and to tangents and normals.
If the derivative rules themselves are the gap, the Fermi Concept Library covers every concept in mathematics, each with its own lesson.
Tonight, redo example 3 for f(x) = x + 4/x and check that the critical points move to x = ±2.
