If ` f (x) = x (sqrtx+sqrt((x+1))`, then

Options

A. f (x) is continuous but not differentiable at x = 0
B. f(x) is differentiable at x = 0
C. f(x) is not differentiable at x = 0
D. None of the above

Explanation

To solve the problem step-by-step, we need to analyze the function \( f(x) = x(\sqrt{x} + \sqrt{x+1}) \) and determine its continuity and differentiability, particularly at \( x = 0 \).

### Step 1: Identify the function
The given function is:
\[ f(x) = x(\sqrt{x} + \sqrt{x+1}) \]

### Step 2: Determine the domain of the function
The function involves square roots, which are defined for non-negative values. Therefore, we need to find the domain where the expression under the square roots is non-negative.

- The term \( \sqrt{x} \) is defined for \( x \geq 0 \).
- The term \( \sqrt{x+1} \) is defined for all \( x \) since \( x + 1 \geq 1 \) for \( x \geq 0 \).

Thus, the domain of \( f(x) \) is:
\[ x \geq 0 \quad \text{or} \quad [0, \infty) \]

### Step 3: Check the value of the function at \( x = 0 \)
Now, we evaluate \( f(0) \):
\[ f(0) = 0(\sqrt{0} + \sqrt{0 + 1}) = 0 \cdot (0 + 1) = 0 \]
So, \( f(0) = 0 \).

### Step 4: Check continuity at \( x = 0 \)
To check if \( f(x) \) is continuous at \( x = 0 \), we need to find the left-hand limit and right-hand limit as \( x \) approaches 0.

- **Left-hand limit**: Since the domain starts from 0, we do not consider left-hand limits for \( x < 0 \) as the function is not defined there.

- **Right-hand limit**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x(\sqrt{x} + \sqrt{x+1}) = 0(\sqrt{0} + \sqrt{1}) = 0 \]

Since \( f(0) = 0 \) and the right-hand limit as \( x \to 0^+ \) is also 0, we conclude:
\[ \lim_{x \to 0} f(x) = f(0) \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 5: Check differentiability at \( x = 0 \)
To check if \( f(x) \) is differentiable at \( x = 0 \), we need to find the derivative \( f'(x) \) and evaluate it at \( x = 0 \).

Using the product rule:
\[ f'(x) = \frac{d}{dx}[x] \cdot (\sqrt{x} + \sqrt{x+1}) + x \cdot \frac{d}{dx}[\sqrt{x} + \sqrt{x+1}] \]
Calculating the derivative:
1. The derivative of \( x \) is \( 1 \).
2. The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \) and for \( \sqrt{x+1} \) is \( \frac{1}{2\sqrt{x+1}} \).

Thus,
\[ f'(x) = (\sqrt{x} + \sqrt{x+1}) + x \left( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x+1}} \right) \]

Evaluating \( f'(0) \):
\[ f'(0) = (\sqrt{0} + \sqrt{1}) + 0 \cdot \left( \frac{1}{2\sqrt{0}} + \frac{1}{2\sqrt{1}} \right) = 1 \]

Since \( f'(0) \) exists, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion
The function \( f(x) \) is continuous and differentiable at \( x = 0 \).