The function `f (x) =cot^(-1) (sqrt((x+3) x)) + cos^(-1) (sqrt(x^2 +3x+1))` is defined for x equal to

Options

A. 0
B. -3
C. `{0,-3}`
D. None

Answer

To find the domain of the function \( f(x) = \cot^{-1}(\sqrt{(x+3)x}) + \cos^{-1}(\sqrt{x^2 + 3x + 1}) \), we need to analyze the conditions under which each component of the function is defined.

### Step 1: Analyze the first term \( \cot^{-1}(\sqrt{(x+3)x}) \)

1. The expression inside the square root, \( (x+3)x \), must be non-negative:
\[ (x+3)x \geq 0 \]
This can be factored as:
\[ x(x + 3) \geq 0 \]

2. To find the roots, we set \( x(x + 3) = 0 \):
- The roots are \( x = 0 \) and \( x = -3 \).

3. We analyze the sign of \( x(x + 3) \) in the intervals determined by the roots:
- For \( x < -3 \): both factors are negative, so the product is positive.
- For \( -3 < x < 0 \): the first factor is negative and the second is positive, so the product is negative.
- For \( x > 0 \): both factors are positive, so the product is positive.

4. Thus, the intervals where \( (x+3)x \geq 0 \) are:
\[ (-\infty, -3] \cup [0, \infty) \]

### Step 2: Analyze the second term \( \cos^{-1}(\sqrt{x^2 + 3x + 1}) \)

1. The expression inside the square root, \( x^2 + 3x + 1 \), must be non-negative:
\[ x^2 + 3x + 1 \geq 0 \]

2. To find the roots, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
Let \( r_1 = \frac{-3 - \sqrt{5}}{2} \) and \( r_2 = \frac{-3 + \sqrt{5}}{2} \).

3. The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), so it is non-negative outside the roots:
\[ (-\infty, r_1] \cup [r_2, \infty) \]

4. Additionally, since we are dealing with \( \cos^{-1} \), the value inside the square root must also be less than or equal to 1:
\[ 0 \leq \sqrt{x^2 + 3x + 1} \leq 1 \]
Squaring gives:
\[ 0 \leq x^2 + 3x + 1 \leq 1 \]
This leads to two inequalities:
- \( x^2 + 3x + 1 \geq 0 \) (already solved).
- \( x^2 + 3x \leq 0 \) simplifies to \( x(x + 3) \leq 0 \), which gives:
\[ [-3, 0] \]

### Step 3: Combine the domains

1. From the first term, we have:
\[ (-\infty, -3] \cup [0, \infty) \]
2. From the second term, we have:
\[ [-3, 0] \]

3. The intersection of these two domains is:
\[ [-3, 0] \]

### Final Domain

Thus, the domain of the function \( f(x) \) is:
\[ \boxed{[-3, 0]} \]