(i) Observe the graph of f'(x) (derivative of the function f(x)) given below and answer the following question

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Q. (i) Observe the graph of f'(x) (derivative of the function f(x)) given below and answer the following questions.

(a) Write the interval in which the function f(x) is decreasing.

(1)

(b) Find the local maximum and local minimum points of the function f(x).

(1)

(ii) Find the absolute maximum and absolute minimum value of the function g(x) = |x| + 2, in the interval [−2, 4]. (2)

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Concept understanding — Derivative Sign Analysis

Derivative Sign Analysis: What the Slope Tells You

Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.

Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.

The Intuition First

If $f'(x)$ is positive, the function is increasing — the graph rises as you move right. If $f'(x)$ is negative, it is decreasing. If $f'(x) = 0$ , there is a horizontal tangent — a potential peak, valley, or flat inflection.

The key: a single point tells you little; you look at intervals. If $f'(x) > 0$ for all $x$ in $(a, b)$ , the function is strictly increasing on that whole interval. Same logic for negative.

Note

The analysis is local — it describes behaviour on intervals, not isolated points. A zero derivative at a single point doesn't guarantee a max or min; check the sign change across that point.

The Precise Statement

Let $f$ be differentiable on an open interval $I$ . Then:

If $f'(x) > 0$ for all $x$ in $I$ , then $f$ is strictly increasing on $I$ .
If $f'(x) < 0$ for all $x$ in $I$ , then $f$ is strictly decreasing on $I$ .
If $f'(x) = 0$ for all $x$ in $I$ , then $f$ is constant on $I$ .

Points where $f'(x) = 0$ (or where $f'$ does not exist) are critical points — the candidates for local maxima and minima.

\text{If } f'(x) > 0 \text{ on } (a,b) \implies f \text{ increasing on } (a,b)

How to Perform It (Step-by-Step)

Find the derivative $f'(x)$ .
Find critical points: solve $f'(x) = 0$ and check where $f'(x)$ is undefined (but $f$ is defined).
Plot these on a number line — they split the domain into intervals.
Pick a test point inside each interval and evaluate $f'$ ; only the sign matters.
Record the sign in each interval and interpret: + means increasing, – means decreasing.

A Concrete Example

Take $f(x) = x^3 - 3x$ .

Step 1: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ .

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

Step 3: Intervals: $(-\infty, -1)$ , $(-1, 1)$ , $(1, \infty)$ .

Step 4: Test points:

$x = -2$ : $f'(-2) = 3(4-1) = 9 > 0$ .
$x = 0$ : $f'(0) = -3 < 0$ .
$x = 2$ : $f'(2) = 9 > 0$ .

Step 5: So $f$ increases on $(-\infty, -1)$ , decreases on $(-1, 1)$ , increases on $(1, \infty)$ . Thus $x = -1$ is a local maximum (sign changes + to –), and $x = 1$ is a local minimum (– to +). …

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Read the sign of f' from its graph to find where f increases/decreases and where it has local extrema; for g(x) = |x| + 2, evaluate at the vertex and both endpoints to find the absolute extrema on a closed interval.

(i) From the graph of $f'(x)$ . The graph shows $f'(x)$ starting positive near $x=-1$ , crossing the x-axis (becoming zero) near $x=1$ , dipping to a minimum near $x=3$ , rising back up and crossing zero again near $x=5$ , then staying positive up to $x=6$ . So $f'(x) > 0$ on $(-1,1)$ , $f'(x) < 0$ on $(1,5)$ , and $f'(x)>0$ again on $(5,6)$ .

Interval where f is decreasing: f decreases exactly where $f'(x)<0$ , i.e. on $\mathbf{(1,5)}$ .
Local maxima/minima of f: At $x=1$ , $f'$ changes sign from positive to negative (f goes from increasing to decreasing), so f has a local maximum at $x=1$ . At $x=5$ , $f'$ changes sign from negative to positive (f goes from decreasing to increasing), so f has a local minimum at $x=5$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set V11 markMCQ
$f(x) = x^2$
(a) Both the Statements I and II are true
(b) Both the Statements I and II are false
(c) Statement I is true and Statement II is false
(d) Statement I is false and Statement II is true
›Reveal solutionSolution

Statement I is false and Statement II is true, so the answer is (d).

Statement I: For $f(x)=x^2$ , $f'(x)=2x$ . On $(0,\infty)$ we have $f'(x)=2x>0$ , so $f$ is increasing there, not decreasing. False. …
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CBSE 2026Set ANNUAL1 markMCQ
$f(x)=\int e^x (x-1)(x-2)\,dx$
(a) $(-\infty,-2)$
(b) $(-2,-1)$
(c) $(1,2)$
(d) $(2,+\infty)$
›Reveal solutionSolution
Since $f(x)=\int e^x(x-1)(x-2)\,dx$ , we get $f'(x)=e^x(x-1)(x-2)$ ; $f$ decreases where $f'(x)<0$ , i.e. on $(1,2)$ .

By the Fundamental Theorem of Calculus, if $f(x)=\int e^x(x-1)(x-2)\,dx$ , then

$f'(x)=e^x(x-1)(x-2)$

A function decreases on an interval where its derivative is negative: $f'(x)<0$ .

Since $e^x>0$ for every real $x$ , the sign of $f'(x)$ is entirely determined by the sign of $(x-1)(x-2)$ :

For $x<1$ : both factors negative $\Rightarrow$ product positive $\Rightarrow f'(x)>0$ . …
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CBSE 2025Set 65/4/11 markMCQ
$\lambda$
›Reveal solutionSolution
A function decreases everywhere when its derivative is non-positive for all $x$ . Here $f'(x) = \cos x + \sin x - \lambda$ must satisfy $\cos x + \sin x \le \lambda$ for all $x$ , which requires $\lambda \ge \sqrt{2}$ (the maximum of $\cos x + \sin x$ ).

A function decreases for all real $x$ when its rate of change is never positive. This translates to the condition $f'(x) \le 0$ for all $x \in \mathbb{R}$ . The question asks us to find which values of the parameter $\lambda$ enforce this condition.

The key insight is that we need to understand the range of the trigonometric expression in the derivative, then choose $\lambda$ large enough to dominate it everywhere.

Finding the derivative

Differentiate $f(x) = \sin x - \cos x - \lambda x + C$ :

$f'(x) = \cos x + \sin x - \lambda$

For $f$ to be decreasing everywhere, we need:

$f'(x) \le 0 \quad \text{for all } x \in \mathbb{R}$
This means:
$\cos x + \sin x - \lambda \le 0$ $\cos x + \sin x \le \lambda \quad \text{for all } x$
Finding the maximum of $\cos x + \sin x$

The condition $\cos x + \sin x \le \lambda$ for all $x$ is equivalent to requiring:

$\lambda \ge \max_{x \in \mathbb{R}} (\cos x + \sin x)$

To find this maximum, we can express the sum as a single sinusoid. Using the identity:

$\cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$
›Proof

Derivation of the identity:

We write $\cos x + \sin x = R \sin(x + \phi)$ for some amplitude $R$ and phase $\phi$ .

Expanding: $R \sin(x + \phi) = R(\sin x \cos \phi + \cos x \sin \phi) = R \cos \phi \cdot \sin x + R \sin \phi \cdot \cos x$

Comparing coefficients:

Coefficient of $\sin x$ : $R \cos \phi = 1$

Coefficient of $\cos x$ : $R \sin \phi = 1$

Squaring and adding: $R^2(\cos^2 \phi + \sin^2 \phi) = 1 + 1 = 2$ , so $R = \sqrt{2}$ .

…
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CBSE 2024Set 65/2/11 markMCQ
$f(x) = x^3 - 3x^2 + 12x - 18$
›Reveal solutionSolution
The derivative $f'(x) = 3x^2 - 6x + 12$ is always positive (its discriminant is negative and leading coefficient positive), so $f(x)$ is strictly increasing on $\mathbb{R}$ . The correct option is (B).

The core question here is about monotonicity — whether a function is always increasing, always decreasing, or neither. For a polynomial, the sign of its derivative tells us everything. If the derivative is positive everywhere, the function is strictly increasing; if negative everywhere, strictly decreasing; if it changes sign, the function is neither.

Let’s see what $f'(x)$ looks like.

Find the derivative. $f(x) = x^3 - 3x^2 + 12x - 18$ Differentiating term by term:

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

Analyze the sign of $f'(x)$ . This is a quadratic: $3x^2 - 6x + 12$ . To check if it ever becomes negative or zero, compute its discriminant:

$D = (-6)^2 - 4 \cdot 3 \cdot 12 = 36 - 144 = -108$
Since $D < 0$ , the quadratic has no real roots — it never touches or crosses the x-axis.

What does a negative discriminant mean for sign? The leading coefficient $3 > 0$ , so the parabola opens upward. A quadratic that opens upward and has no real roots is always positive. Therefore, $f'(x) > 0$ for every real $x$ . …
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Q.(i) Observe the graph of f'(x) (derivative of the function f(x)) given below and answer the following questions.

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