Match List-I with List-II

The function $f(x) = (x − 1)(x + 1)^2$ has

Choose the correct answer from the options given below:

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)

The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)

Given: $f(x) = (x - 1)(x + 1)^2$

Compute first derivative:

$f'(x) = \frac{d}{dx}[(x - 1)(x + 1)^2]$

$= (x - 1) \cdot \frac{d}{dx}[(x + 1)^2] + (x + 1)^2 \cdot \frac{d}{dx}[x - 1]$

$= (x - 1)(2)(x + 1) + (x + 1)^2(1)$

$= 2(x - 1)(x + 1) + (x + 1)^2$

$= (x + 1)[2(x - 1) + (x + 1)]$

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

Critical points from $f'(x) = 0$:

$(x + 1)(3x - 1) = 0 \Rightarrow x = -1,\ \frac{1}{3}$

Compute second derivative:

$f''(x) = \frac{d}{dx}[(x + 1)(3x - 1)] = (x + 1)(3) + (3x - 1)(1)$

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

At $x = -1$: $f''(-1) = 6(-1) + 2 = -4$ ⟹ local maxima

At $x = \frac{1}{3}$: $f''\left(\frac{1}{3}\right) = 6 \cdot \frac{1}{3} + 2 = 4$ ⟹ local minima

Compute $f(-1) = (-1 - 1)(-1 + 1)^2 = (-2)(0)^2 = 0$

Compute $f\left(\frac{1}{3}\right) = \left(\frac{1}{3} - 1\right)\left(\frac{1}{3} + 1\right)^2 = \left(-\frac{2}{3}\right)\left(\frac{4}{3}\right)^2 = -\frac{2}{3} \cdot \frac{16}{9} = -\frac{32}{27}$