Which of the following statement is true for the function `f(x)={{:(sqrt(x),","x ge 1),(x^(3) ,","0 le x lt 1),((x^(3))/(3)-4x,"," x lt 0):}`

Options

A. It is monotonic increasing `AA x in R `
B. f'(x) fails to exist for 3 distinct real values of x
C. f'(x) changes its sign twice as x varies from `(-oo,oo)`
D. Function attains its extreme values at `x_(1) and x_(2)` , such that ` x_(1)x_(2) gt 0`

Answer

To determine which statement is true for the given piecewise function \( f(x) \), we need to analyze the function and its derivative step by step.

### Step 1: Define the function
The function \( f(x) \) is defined as follows:
- \( f(x) = \sqrt{x} \) for \( x \geq 1 \)
- \( f(x) = x^3 \) for \( 0 \leq x < 1 \)
- \( f(x) = \frac{x^3}{3} - 4x \) for \( x < 0 \)

### Step 2: Find the derivative \( f'(x) \)
We will find the derivative for each piece of the function.

1. For \( x \geq 1 \):
\[ f'(x) = \frac{1}{2\sqrt{x}} \]

2. For \( 0 \leq x < 1 \):
\[ f'(x) = 3x^2 \]

3. For \( x < 0 \):
\[ f'(x) = x^2 - 4 \]

### Step 3: Find critical points
Critical points occur where \( f'(x) = 0 \) or is undefined.

1. For \( x \geq 1 \):
\[ f'(x) \text{ is never } 0 \text{ (always positive)} \]

2. For \( 0 \leq x < 1 \):
\[ 3x^2 = 0 \implies x = 0 \]

3. For \( x < 0 \):
\[ x^2 - 4 = 0 \implies x = \pm 2 \implies x = -2 \text{ (since we only consider } x < 0\text{)} \]

### Step 4: Analyze the sign of \( f'(x) \)
We will analyze the sign of \( f'(x) \) around the critical points \( x = -2 \) and \( x = 0 \).

- For \( x < -2 \):
\[ f'(-3) = (-3)^2 - 4 = 9 - 4 = 5 \quad (\text{positive}) \]

- For \( -2 < x < 0 \):
\[ f'(-1) = (-1)^2 - 4 = 1 - 4 = -3 \quad (\text{negative}) \]

- For \( 0 < x < 1 \):
\[ f'(0.5) = 3(0.5)^2 = 3 \cdot 0.25 = 0.75 \quad (\text{positive}) \]

- For \( x \geq 1 \):
\[ f'(1) = \frac{1}{2\sqrt{1}} = 0.5 \quad (\text{positive}) \]

### Step 5: Conclusion on monotonicity
From the sign analysis:
- \( f'(x) \) changes from positive to negative at \( x = -2 \) (local maximum).
- \( f'(x) \) changes from negative to positive at \( x = 0 \) (local minimum).
- Therefore, \( f(x) \) is not monotonic over its entire domain.

### Step 6: Evaluate the statements
1. **Option 1**: It is a monotonic increasing function. **(False)**
2. **Option 2**: \( f'(x) \) fails to exist for three distinct real values of \( x \). **(False)**
3. **Option 3**: \( f'(x) \) changes its sign twice as \( x \) varies from \( -\infty \) to \( +\infty \). **(True)**
4. **Option 4**: The function attains its extreme values at \( x_1 \) and \( x_2 \) such that \( x_1 \cdot x_2 > 0 \). **(False)**

### Final Answer
The correct option is **Option 3**.