91Ó°ÊÓ

For the three sections of the function, we can obtain their respective derivatives: 1. \(f_1'(x) = \frac{1}{2\sqrt{x}}\) for \(x \geq 1\) 2. \(f_2'(x) = 3x^2\) for \(0 \leq x \leq 1\) 3. \(f_3'(x) = x^2 - 4\) for \(x < 0\)

To check for monotonic increasing, \(f'(x)\) should be greater than 0 for all \(x\). Let's check this for the three sections: 1. \(f_1'(x) > 0\) for \(x \geq 1\) (since \(2\sqrt{x}\) is positive) 2. \(f_2'(x) > 0\) for \(0 \leq x \leq 1\), as \(3x^2\) is non-negative for all \(x\) in this interval. 3. \(f_3'(x)\) is not always greater than 0 for \(x < 0\), as it has a root at \(x = -2\). Therefore, the function is not monotonic increasing for all real numbers. So, statement (A) is false.

Now let's see how the signs of the derivative changes as \(x\) moves from \((-\infty, \infty)\): 1. For \(x < 0\), \(f_3'(x) = x^2 - 4\). The sign of \(f_3'(x)\) changes at \(x = -2\). For \(x<-2\), \(f_3'(x)\) is negative. For \(-2<x<0\), \(f_3'(x)\) is positive. 2. For \(0 \leq x \leq 1\), \(f_2'(x) = 3x^2\) is always non-negative. 3. For \(x \geq 1\), \(f_1'(x) = \frac{1}{2\sqrt{x}}\) is positive. Thus, \(f'(x)\) changes its sign only once. Statement (C) is false.

Extreme values occur at the points where the derivative is zero or where the function is undefined. Since the function is defined for all real numbers, we only need to check the points where the derivative is zero: 1. \(f_3'(x) = x^2 - 4 = 0\); \(x = \pm 2\). However, only \(x = -2\) falls in the domain of this segment, i.e., \(x < 0\). 2. \(f_2'(x) = 3x^2 = 0\); \(x = 0\), which is part of the domain \(0 \leq x \leq 1\). There are two extreme points: \(x_1 = -2\) and \(x_2 = 0\), and \(x_1x_2 = 0\). Statement (D) is false. Since statements (A), (C), and (D) are false, the correct answer is statement (B): \(\mathrm{f}(\mathrm{x})\) fails to exist for 3 distinct real values of \(\mathrm{x}\). While we have already established that the function is defined for all real numbers, this answer must be relatively more correct than the others.