Increasing and decreasing functions

Look at any smooth curve and ask a simple question: as you slide left to right, is the graph climbing or falling? A tangent line answers it instantly — where the tangent tilts upward the curve is rising, where it tilts downward the curve is falling, and the slope of that tangent is exactly the derivative $f'(x)$ . This turns a visual, hand-wavy question about a picture into a precise algebraic test: read the sign of $f'(x)$ and you know, without ever sketching the curve, whether the function is increasing or decreasing on that stretch. This article builds that sign test from the definition, applies it step by step to a cubic polynomial to map out its full increasing/decreasing behaviour, and flags the traps — like assuming a function is monotonic everywhere — that cost the most marks. The same first-derivative sign analysis is the tool you reuse immediately in the next topic, maxima and minima.

The core insight

A differentiable function is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$ , so locating the zeros of $f'(x)$ and testing the sign in between splits the domain into intervals of increase and decrease.

The core idea

Let $f$ be differentiable on an interval $I$ .

$f$ is increasing on $I$ if $f'(x) \geq 0$ for every $x \in I$ (strictly increasing if $f'(x) > 0$ throughout the interior, except possibly at isolated points).
$f$ is decreasing on $I$ if $f'(x) \leq 0$ for every $x \in I$ (strictly decreasing if $f'(x) < 0$ throughout the interior, except possibly at isolated points).

Geometrically, a positive derivative means the tangent has positive slope, so the curve rises as $x$ increases; a negative derivative means it falls. The CBSE syllabus states and uses this as a motivated fact — the formal justification (the Mean Value Theorem) sits outside the current syllabus, so no proof is expected.

f'(x) > 0 \ \Rightarrow\ f \text{ increasing}, \qquad f'(x) < 0 \ \Rightarrow\ f \text{ decreasing}. \qquad (1)

where: $f'(x)$ is the first derivative of $f$ at $x$ , and "increasing/decreasing" describes the behaviour of $f$ on the interval around that point, not at a single isolated $x$ -value. The working method has four moves: differentiate $f$ ; solve $f'(x) = 0$ to find the critical points, which split the domain into subintervals; pick a convenient test value inside each subinterval and check the sign of $f'(x)$ there; then read off increasing wherever $f'(x) > 0$ and decreasing wherever $f'(x) < 0$ . The picture below shows exactly this for a cubic with two critical points. It helps to keep the two readings of $f'(x)$ straight: as a number, $f'(x)$ is the instantaneous rate of change of $f$ at that point; as a slope, it is the steepness and direction of the tangent line drawn at that point on the graph. Both readings agree perfectly — a large positive $f'(x)$ means $f$ is rising steeply, a small positive $f'(x)$ means it is rising gently, and $f'(x) = 0$ means the tangent is momentarily flat, which is exactly why critical points are the natural places to cut the domain into pieces.

Figure

Number line for a cubic derivative showing plus sign then minus sign then plus sign across critical points 2 and 3

Derivation ABuilding the sign-of-derivative test into intervals

The goal is to convert the raw definitions of increasing and decreasing into a concrete, repeatable four-step recipe that works for any differentiable function whose derivative is a polynomial, so that a sign chart can be built instead of relying on a sketch of the graph.

Step 1. Differentiate the given function to get $f'(x)$ . This is possible because CBSE-level increasing/decreasing questions always start from a differentiable $f$ , usually a polynomial or a simple rational or trigonometric expression.

Step 2. Solve $f'(x) = 0$ to find the critical points. Because $f'$ is continuous wherever $f$ is differentiable, it can only switch sign at a zero of $f'$ (or at a point where $f'$ is undefined, for rational functions). These points are exactly where the domain must be cut into subintervals.

Step 3. In each subinterval created by the critical points, the sign of $f'(x)$ cannot change — pick any one convenient test point in the subinterval and evaluate $f'$ there; that single sign holds for the whole subinterval. This works because $f'$ itself is continuous on each subinterval and never touches zero inside it (all its zeros were already removed as the boundary points), so by continuity it cannot flip from positive to negative without passing through zero somewhere in between — one test point is therefore enough to fix the sign for the entire piece, no matter how large the piece is.

Step 4. Translate signs into behaviour using the definitions: $f'(x) > 0$ on a subinterval means $f$ is (strictly) increasing there, and $f'(x) < 0$ means $f$ is (strictly) decreasing there. Stitching the subintervals together gives the complete increasing/decreasing description of $f$ over its whole domain.

Worked example

Find the intervals in which $f(x) = 2x^3 - 15x^2 + 36x + 1$ is increasing and decreasing.

Step 1. Differentiate: $f'(x) = 6x^2 - 30x + 36$ . Factor out $6$ and factor the quadratic: $f'(x) = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$ .

Step 2. Set $f'(x) = 0$ : the critical points are $x = 2$ and $x = 3$ . These split $\mathbb{R}$ into three subintervals: $(-\infty, 2)$ , $(2, 3)$ , and $(3, \infty)$ .

Step 3. Test each subinterval. On $(-\infty, 2)$ , take $x = 0$ : $f'(0) = 6(-2)(-3) = 36 > 0$ . On $(2, 3)$ , take $x = 2.5$ : $f'(2.5) = 6(0.5)(-0.5) = -1.5 < 0$ . On $(3, \infty)$ , take $x = 4$ : $f'(4) = 6(2)(1) = 12 > 0$ .

Step 4. Read off the behaviour. $f$ is increasing on $(-\infty, 2) \cup (3, \infty)$ and decreasing on $(2, 3)$ .

Competitive stretch

Show that $f(x) = x^2 - x + 1$ is neither increasing nor decreasing on $(-1, 1)$ , and identify where it is strictly increasing and strictly decreasing on $\mathbb{R}$ .

Step 1. Differentiate: $f'(x) = 2x - 1$ , giving the single critical point $x = \tfrac{1}{2}$ .

Step 2. For $x < \tfrac12$ , $f'(x) < 0$ , so $f$ is strictly decreasing on $\left(-\infty, \tfrac12\right)$ ; for $x > \tfrac12$ , $f'(x) > 0$ , so $f$ is strictly increasing on $\left(\tfrac12, \infty\right)$ .

Step 3. Since $\tfrac12$ lies inside $(-1, 1)$ , the derivative is negative for the left part of the interval and positive for the right part. Hence $f$ is neither increasing nor decreasing on the whole interval $(-1, 1)$ — monotonicity claims must always be checked on intervals lying entirely on one side of a critical point.

Common traps

Testing the sign of $f(x)$ instead of $f'(x)$ . Increasing/decreasing behaviour is governed by the derivative's sign, never the function's own value.
Assuming a function is monotonic on its whole domain. Always find every critical point first — a function can rise on part of an interval and fall on another part, exactly as in the Competitive stretch example above.
Sloppy factoring of $f'(x)$ . A wrongly factored quadratic derivative gives wrong critical points and hence a wrong sign chart. Expand the factors back out to double-check.
Ignoring points where $f'(x)$ is undefined. For rational or root functions, points where the derivative fails to exist can also split the domain into monotonicity intervals, just like the zeros of $f'(x)$ .

Ask the tutor: Want a fresh practice problem with a trigonometric or rational $f(x)$ and a full sign-chart walkthrough? Ask the AI tutor for one.

Do I need to prove why f prime greater than zero means f is increasing?

No. CBSE Class 12 treats this as a stated, motivated fact you apply directly — the rigorous justification uses the Mean Value Theorem, which is outside the current syllabus. In an exam you differentiate, find critical points, build the sign chart, and state the conclusion; you are not asked to prove the underlying theorem.

What if f prime x equals zero over a whole stretch, not just at isolated points?

Then $f$ is constant on that stretch, not strictly increasing or decreasing there, though it still counts as increasing (and decreasing) in the non-strict, weak sense since $f'(x) \geq 0$ and $f'(x) \leq 0$ both hold. At the board level almost every problem gives a polynomial whose derivative is zero only at isolated points, so this subtlety rarely changes the final answer, but it is worth naming if a question explicitly asks about strict monotonicity.

How is this different from finding local maxima and minima?

It is the same sign-chart machinery applied to a different question. Here you read off where $f$ rises and falls; in maxima and minima you look at the specific points where $f'(x)$ changes sign — a switch from positive to negative marks a local maximum, and negative to positive marks a local minimum. Master the sign chart here and the next topic is largely relabelling the same critical points.

Increasing and decreasing functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads