Find the intervals in which the function f given b – Class 12 Mathematics

Question of 188

Q. Find the intervals in which the function f given by f(x) = 4x^3 - 6x^2 - 72x + 30 is:

(a) Strictly increasing

(b) Strictly decreasing OR The volume of a cube is increasing at the rate of 9 cm^3/sec. If its core is 10 cm at what rate the surface area is increasing? [Note: source paper prints "core"; likely a printing typo for "edge", transcribed verbatim as printed.]

Chhattisgarh CgbseCGBSE Intermediate Board 2023Subjective· 4mImportance★★★★★

0% · 0/188 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Monotonic Function Analysis

Monotonic Function Analysis

A function is monotonic on an interval when it moves in a single direction across that interval — either always rising or always falling, never doubling back. Derivatives give us a clean, mechanical way to detect this, which is why monotonicity is one of the first applications of the derivative.

Increasing, Decreasing, Monotonic

On an interval $I$ , a function $f$ is:

increasing if $x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$ ,
strictly increasing if $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ ,
decreasing if $x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2)$ ,
strictly decreasing if $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$ .

A function that is either increasing throughout $I$ or decreasing throughout $I$ is called monotonic on $I$ .

The Derivative Test

The slope of the tangent tells you the direction of travel. If $f$ is differentiable on an open interval $I$ :

f'(x) > 0 \text{ on } I \implies f \text{ is strictly increasing on } I

The idea is intuitive: a positive slope means the graph climbs as you move right, a negative slope means it falls.

How to Analyse Monotonicity

Compute $f'(x)$ .
Solve $f'(x) = 0$ (and note where $f'$ is undefined). These critical points split the domain into intervals.
Test the sign of $f'$ in each interval.
Read off where $f$ increases ( $f' > 0$ ) and decreases ( $f' < 0$ ).

Example. For $f(x) = x^2 - 4x + 1$ , $f'(x) = 2x - 4$ . So $f'(x) < 0$ for $x < 2$ and $f'(x) > 0$ for $x > 2$ : the function decreases on $(-\infty, 2)$ and increases on $(2, \infty)$ . …

Start your 14-day free trial to unlock the full solution →

Find $f'(x)$ , factor it, and read off the sign in each interval determined by its roots.

Given $f(x) = 4x^3-6x^2-72x+30$ .

Step 1 — differentiate.

$f'(x) = 12x^2-12x-72 = 12(x^2-x-6) = 12(x-3)(x+2)$

Step 2 — find critical points. $f'(x)=0$ at $x=-2$ and $x=3$ . These divide the real line into three intervals: $(-\infty,-2)$ , $(-2,3)$ , $(3,\infty)$ .

Step 3 — test the sign of $f'(x)=12(x-3)(x+2)$ in each interval:

For $x<-2$ (e.g. $x=-3$ ): $(x-3)<0$ , $(x+2)<0$ , product positive $\Rightarrow f'(x)>0$ .
For $-2<x<3$ (e.g. $x=0$ ): $(x-3)<0$ , $(x+2)>0$ , product negative $\Rightarrow f'(x)<0$ .
For $x>3$ (e.g. $x=4$ ): both factors positive $\Rightarrow f'(x)>0$ .

So $f$ is strictly increasing on $(-\infty,-2)\cup(3,\infty)$ and strictly decreasing on $(-2,3)$ .

OR — related rates. A cube of edge $x$ has volume $V=x^3$ and surface area $S=6x^2$ . Given $\dfrac{dV}{dt}=9\ \text{cm}^3/\text{s}$ , find $\dfrac{dS}{dt}$ when $x=10$ cm.

…

Start your 14-day free trial to unlock the full solution →

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

CBSE 2024Set ANNUAL1 markMCQ
$Q. The interval in which the function f(x) = [ILLEGIBLE — page defect: heavy scan grain/noise obscures the printed function and range], 0 \leq x \leq [ILLEGIBLE] is strictly [ILLEGIBLE — increasing or decreasing, not legible] :$
(a) [ILLEGIBLE — page defect]
(b) [ILLEGIBLE — page defect]
(c) [ILLEGIBLE — page defect]
(d) [ILLEGIBLE — page defect]
›Reveal solutionSolution

This question cannot be honestly answered: the printed function, its domain range, whether it asks for increasing or decreasing, and all four options are lost to a heavy scan-grain/noise defect on the source PDF.

Note on the source scan: the question stem (which function f(x), the exact interval endpoint(s), and whether it asks about strictly increasing or strictly decreasing) and all four multiple-choice options are obscured by heavy scan grain/noise on the original page. Rather than invent a plausible-looking function and answer, we report this honestly.

…

Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 mark
$(-1, 1)$
›Reveal solutionSolution

The sign of $f'(x)$ changes at $x=\tfrac12$ , which lies inside $(-1,1)$ , so $f$ is not monotonic on the whole interval.

Given $f(x)=x^2-x+1$ , so $f'(x)=2x-1$ .

Set $f'(x)=0$ : $2x-1=0 \Rightarrow x=\tfrac12$ , which lies inside $(-1,1)$ .

For $-1<x<\tfrac12$ : $f'(x)=2x-1<0$ , so $f$ is decreasing on $\left(-1,\tfrac12\right)$ .

For $\tfrac12<x<1$ : $f'(x)=2x-1>0$ , so $f$ is increasing on $\left(\tfrac12,1\right)$ .

…

Start your 14-day free trial to see the full solution →

←Previous chapterCh 5Continuity and Differentiability

Q.Find the intervals in which the function f given by f(x) = 4x^3 - 6x^2 - 72x + 30 is:

🎓Unlock everything free for 14 days

Unlock everything free for 14 days