Increasing and decreasing functions — first derivative test interpretation

Question

How do we determine whether a function is increasing or decreasing on an interval? Find the intervals on which $f(x) = 2x^3 - 9x^2 + 12x - 5$ is increasing or decreasing.

Solution — Step by Step

A function $f(x)$ is:

Increasing on an interval if $f'(x) > 0$ for all $x$ in that interval
Decreasing on an interval if $f'(x) < 0$ for all $x$ in that interval
Constant if $f'(x) = 0$ throughout

We find the critical points (where $f'(x) = 0$ or undefined), then check the sign of $f'(x)$ in each interval.

f(x) = 2x^3 - 9x^2 + 12x - 5

f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)

Critical points: $x = 1$ and $x = 2$ .

Interval	Test value	$f'(x) = 6(x-1)(x-2)$	Sign	Behaviour
$(-\infty, 1)$	$x = 0$	$6(0-1)(0-2) = 6(−1)(−2) = 12$	Positive	Increasing
$(1, 2)$	$x = 1.5$	$6(0.5)(−0.5) = -1.5$	Negative	Decreasing
$(2, \infty)$	$x = 3$	$6(2)(1) = 12$	Positive	Increasing

$f(x)$ is increasing on $(-\infty, 1) \cup (2, \infty)$ and decreasing on $(1, 2)$ .

At $x = 1$ , $f$ has a local maximum. At $x = 2$ , $f$ has a local minimum.

flowchart TD
    A[Given f of x] --> B[Find f prime of x]
    B --> C[Set f prime = 0, find critical points]
    C --> D[Divide number line into intervals]
    D --> E[Test sign of f prime in each interval]
    E --> F{f prime positive?}
    F -->|Yes| G[Increasing on that interval]
    F -->|No| H[Decreasing on that interval]

Why This Works

The derivative $f'(x)$ represents the slope of the tangent at any point. Positive slope means the function is going up (increasing). Negative slope means going down (decreasing). The critical points where $f'(x) = 0$ are where the function changes direction (local maxima or minima).

Alternative Method

For JEE: we can also check monotonicity by evaluating $f'(x)$ directly as a quadratic. Since $f'(x) = 6(x-1)(x-2)$ is a quadratic opening upward (positive leading coefficient), it is negative between its roots (1, 2) and positive outside. No need for a sign chart — just use the parabola’s shape.

Common Mistake

Students sometimes include the critical points in the “increasing” or “decreasing” intervals. Strictly, $f'(x) = 0$ at the critical points — the function is neither increasing nor decreasing at those exact points. In CBSE, you can use closed brackets $[1, 2]$ , but in JEE, the convention is open intervals $(1, 2)$ for the decreasing region.

For JEE-level problems, once you have $f'(x)$ factored, quickly sketch the sign of $f'(x)$ : positive outside the roots, negative between them (if leading coefficient is positive). If the leading coefficient is negative, it is the opposite. This mental shortcut saves time.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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