Match List-I with List-II

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

List-I	List-II
(A) A local maxima at $x =$ ____	(I) $\frac{1}{3}$
(B) A local minima at $x =$ ____	(II) 0
(C) The local minimum value of $f(x) =$ ____	(III) -1
(D) The local maximum value of $f(x) =$ ____	(IV) $-\frac{32}{27}$

Choose the correct answer from the options given below:

The function $f(x) = (x - 1)(x + 1)^2$ needs to be analyzed for local maxima and minima.

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Finding the derivative with $u = (x-1)$ and $v = (x+1)^2$:

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

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

Factoring out $(x+1)$:

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

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

$f'(x) = (x+1)(3x - 1)$

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Setting $f'(x) = 0$:

$(x+1)(3x-1) = 0$

Critical points: $x = -1$ or $x = \frac{1}{3}$

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Finding $f''(x)$:

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

$f''(x) = 6x + 2$

At $x = -1$:

$f''(-1) = 6(-1) + 2 = -4 < 0$

Local maximum at $x = -1$

At $x = \frac{1}{3}$:

$f''(\frac{1}{3}) = 6(\frac{1}{3}) + 2 = 4 > 0$

Local minimum at $x = \frac{1}{3}$

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At $x = -1$:

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

$= (-2)(0)^2$

$= 0$

At $x = \frac{1}{3}$:

$f(\frac{1}{3}) = (\frac{1}{3}-1)(\frac{1}{3}+1)^2$

$= (-\frac{2}{3})(\frac{4}{3})^2$

$= (-\frac{2}{3}) \times \frac{16}{9}$

$= -\frac{32}{27}$

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(A) Local maxima at $x = -1$ → (III)

(B) Local minima at $x = \frac{1}{3}$ → (I)

(C) Local minimum value $= -\frac{32}{27}$ → (IV)

(D) Local maximum value $= 0$ → (II)

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