Range of `f(x)=sin^(-1)x+x^(2)+4x+1` is :

Options

A. `[-(pi)/(2)-2, (pi)/(2)+6]`
B. `[0, (pi)/(2)+6]`
C. `[-(pi)/(2)-2, oo)`
D. `(-3, oo)`

Answer

To find the range of the function \( f(x) = \sin^{-1}(x) + x^2 + 4x + 1 \), we will analyze the components of the function step by step.

### Step 1: Determine the domain of \( \sin^{-1}(x) \)
The function \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, the domain of \( f(x) \) is also \([-1, 1]\).

**Hint:** Remember that the inverse sine function has a restricted domain and range.

### Step 2: Find the maximum and minimum values of \( f(x) \)
We need to evaluate \( f(x) \) at the endpoints of the interval \([-1, 1]\).

1. **Evaluate at \( x = 1 \)**:
\[ f(1) = \sin^{-1}(1) + 1^2 + 4 \cdot 1 + 1 \]
\[ = \frac{\pi}{2} + 1 + 4 + 1 = \frac{\pi}{2} + 6 \]

2. **Evaluate at \( x = -1 \)**:
\[ f(-1) = \sin^{-1}(-1) + (-1)^2 + 4 \cdot (-1) + 1 \]
\[ = -\frac{\pi}{2} + 1 - 4 + 1 = -\frac{\pi}{2} - 2 \]

### Step 3: Determine the range of \( f(x) \)
Now that we have the values of \( f(x) \) at the endpoints:
- Maximum value at \( x = 1 \): \( \frac{\pi}{2} + 6 \)
- Minimum value at \( x = -1 \): \( -\frac{\pi}{2} - 2 \)

Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{\pi}{2} - 2, \frac{\pi}{2} + 6\right] \]

### Final Answer
The range of \( f(x) = \sin^{-1}(x) + x^2 + 4x + 1 \) is:
\[ \left[-\frac{\pi}{2} - 2, \frac{\pi}{2} + 6\right] \]