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Q6(e):
Find the intervals in which the following functions are strictly increasing or decreasing: (e) $(x + 1)^3 (x – 3)^3$

Solution :

Given Function and Initial Setup

The function is given by $f(x) = (x + 1)^3 (x - 3)^3$.

To determine the intervals of monotonicity, we must find the first derivative $f'(x)$ and determine its sign across the domain of the function.

Step 1: Differentiation using the Product Rule

Let $f(x) = u(x)v(x)$, where $u(x) = (x + 1)^3$ and $v(x) = (x - 3)^3$.

[By the Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$]

$f'(x) = 3(x + 1)^2 (x - 3)^3 + (x + 1)^3 \cdot 3(x - 3)^2$

Factor out the common terms $3(x + 1)^2 (x - 3)^2$:

$f'(x) = 3(x + 1)^2 (x - 3)^2 [(x - 3) + (x + 1)]$

$f'(x) = 3(x + 1)^2 (x - 3)^2 [2x - 2]$

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

Step 2: Determining Critical Points

To find the critical points, set $f'(x) = 0$:

$6(x + 1)^2 (x - 3)^2 (x - 1) = 0$

The critical points are $x = -1$, $x = 1$, and $x = 3$.

Step 3: Interval Analysis

The critical points divide the real number line into four intervals: $(-\infty, -1)$, $(-1, 1)$, $(1, 3)$, and $(3, \infty)$.

Interval Test Point ($x$) Sign of $f'(x) = 6(x + 1)^2 (x - 3)^2 (x - 1)$ Nature of Function
$(-\infty, -1)$ $-2$ $6(+)(+)(-3) < 0$ Strictly Decreasing
$(-1, 1)$ $0$ $6(+)(+)(-1) < 0$ Strictly Decreasing
$(1, 3)$ $2$ $6(+)(+)(1) > 0$ Strictly Increasing
$(3, \infty)$ $4$ $6(+)(+)(3) > 0$ Strictly Increasing

Step 4: Synthesis of Intervals

Since $(x+1)^2$ and $(x-3)^2$ are always non-negative for all $x \in \mathbb{R}$, the sign of $f'(x)$ is governed primarily by $(x-1)$.

  • $f'(x) < 0$ for $x \in (-\infty, -1) \cup (-1, 1)$. Since the function is continuous at $x = -1$, the interval of decrease is $(-\infty, 1)$.
  • $f'(x) > 0$ for $x \in (1, 3) \cup (3, \infty)$. Since the function is continuous at $x = 3$, the interval of increase is $(1, \infty)$.

Visual Representation of $f'(x)$ Sign Change

-1 1 3 f'(x) < 0 f'(x) < 0 f'(x) > 0 f'(x) > 0

Final Solution: The function is strictly decreasing on $(-\infty, 1]$ and strictly increasing on $[1, \infty)$.


More Questions from Class 12 Mathematics Application of Derivatives EXERCISE 6.2


CBSE Solutions for Class 12 Mathematics Application of Derivatives


Chapters in CBSE - Class 12 Mathematics


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