Match List-I with List-II

List-I	List-II
Function f(x)	Interval for increasing/decreasing of function f(x)
---	---
(A) $f(x) = x\vert x\vert$	(I) Decreases on $(0, \infty)$
(B) $f(x) = x^2 + 2x - 5$	(II) Increases on $(3, \infty)$
(C) $f(x) = x^2 - 6x + 9$	(III) Decreases on $(-\infty, -1)$
(D) $f(x) = -x^2$	(IV) Increases on $(-\infty, \infty)$

Choose the correct answer from the options given below:

If $f'(x) > 0$ on an interval, the function is increasing there. If $f'(x) < 0$, it is decreasing.

$f(x) = x|x|$

$f(x) = \begin{cases} x^2, & \text{if } x \geq 0 \\ -x^2, & \text{if } x < 0 \end{cases}$

$f'(x) = \begin{cases} 2x \geq 0, & \text{when } x \geq 0 \\ -2x > 0, & \text{when } x < 0 \end{cases}$

$f'(x) \geq 0$ everywhere, so $f(x)$ is increasing on $(-\infty, \infty)$

$\Rightarrow$ (A) → (IV)

$f(x) = x^2 + 2x - 5$

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

For $x < -1$, $f'(x) < 0$, so $f(x)$ is decreasing on $(-\infty, -1)$

$\Rightarrow$ (B) → (III)

$f(x) = x^2 - 6x + 9 = (x - 3)^2$

$f'(x) = 2x - 6 = 2(x - 3)$

For $x > 3$, $f'(x) > 0$, so $f(x)$ is increasing on $(3, \infty)$

$\Rightarrow$ (C) → (II)

$f(x) = -x^2$

$f'(x) = -2x$

For $x > 0$, $f'(x) < 0$, so $f(x)$ is decreasing on $(0, \infty)$

$\Rightarrow$ (D) → (I)

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