Increasing and Decreasing Functions — First Derivative Test Interpretation

medium CBSE JEE-MAIN 3 min read

Question

How do we determine the intervals where a function is increasing or decreasing using the first derivative test?


Solution — Step by Step

For a differentiable function f(x)f(x) on an interval:

  • f(x)>0f'(x) > 0 for all xx in the interval     \implies ff is strictly increasing
  • f(x)<0f'(x) < 0 for all xx in the interval     \implies ff is strictly decreasing
  • f(x)=0f'(x) = 0 for all xx in the interval     \implies ff is constant

The derivative tells us the “direction” of the function at each point.

Set f(x)=0f'(x) = 0 and solve. The solutions are critical points — these are where the function might switch from increasing to decreasing (or vice versa).

Example: f(x)=x36x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1

f(x)=3x212x+9=3(x24x+3)=3(x1)(x3)f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Critical points: x=1x = 1 and x=3x = 3.

The critical points divide the number line into intervals. Pick a test point in each:

IntervalTest pointf(x)=3(x1)(x3)f'(x) = 3(x-1)(x-3)SignBehaviour
(,1)(-\infty, 1)x=0x = 03(01)(03)=93(0-1)(0-3) = 9++Increasing
(1,3)(1, 3)x=2x = 23(21)(23)=33(2-1)(2-3) = -3-Decreasing
(3,)(3, \infty)x=4x = 43(41)(43)=93(4-1)(4-3) = 9++Increasing

So ff is increasing on (,1)(3,)(-\infty, 1) \cup (3, \infty) and decreasing on (1,3)(1, 3).

graph TD
    A[Given f of x] --> B[Compute f prime of x]
    B --> C[Solve f prime of x = 0 for critical points]
    C --> D[Divide number line into intervals]
    D --> E[Pick test point in each interval]
    E --> F[Evaluate sign of f prime at test point]
    F --> G{f prime positive?}
    G -->|Yes| H[Function increasing on that interval]
    G -->|No| I[Function decreasing on that interval]

Why This Works

The derivative f(x)f'(x) measures the slope of the curve at every point. A positive slope means the curve goes upward (increasing), and a negative slope means it goes downward (decreasing). Critical points are where the slope is zero — the curve is momentarily flat — and these are the only places where the behaviour can switch.

CBSE 12 boards and JEE Main both test this heavily. The standard question format is: “Find the intervals of increase and decrease for f(x)f(x).” Full marks require showing the sign chart with test points. Do not just state the intervals without justification.


Alternative Method

Instead of testing individual points, you can use the factor analysis directly. For f(x)=3(x1)(x3)f'(x) = 3(x-1)(x-3):

Each factor changes sign at its root. Track signs:

  • (x1)(x-1): negative for x<1x < 1, positive for x>1x > 1
  • (x3)(x-3): negative for x<3x < 3, positive for x>3x > 3

Multiply the signs in each interval to get the overall sign of f(x)f'(x).


Common Mistake

Students write “increasing on (,1](-\infty, 1] and [3,)[3, \infty)” with square brackets at the critical points. Strictly speaking, f(1)=0f'(1) = 0 means the function is neither increasing nor decreasing at that exact point. CBSE boards accept both open and closed intervals, but for JEE, use open intervals for strict monotonicity: (,1)(-\infty, 1) and (3,)(3, \infty).

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