If `f(x)=(sin^2 x+4sinx+5)/(2sin^2x+8sinx+8)` then range of `f(x)` is

Explanation

To find the range of the function \( f(x) = \frac{\sin^2 x + 4 \sin x + 5}{2 \sin^2 x + 8 \sin x + 8} \), we can follow these steps:

### Step 1: Rewrite the Function
The function can be rewritten by factoring out constants from the numerator and the denominator:
\[ f(x) = \frac{\sin^2 x + 4 \sin x + 5}{2(\sin^2 x + 4 \sin x + 4)} \]
This simplifies to:
\[ f(x) = \frac{1}{2} \cdot \frac{\sin^2 x + 4 \sin x + 5}{\sin^2 x + 4 \sin x + 4} \]

### Step 2: Identify the Components
Let \( y = \sin x \). Then we can express the function in terms of \( y \):
\[ f(y) = \frac{y^2 + 4y + 5}{y^2 + 4y + 4} \]

### Step 3: Simplify the Function
Notice that the denominator can be factored:
\[ f(y) = \frac{y^2 + 4y + 5}{(y + 2)^2} \]
This means we can rewrite the function as:
\[ f(y) = \frac{1}{2} + \frac{1}{(y + 2)^2} \]

### Step 4: Determine the Range of \( y \)
Since \( y = \sin x \), the range of \( y \) is:
\[ -1 \leq y \leq 1 \]

### Step 5: Analyze the Function Behavior
To find the minimum and maximum values of \( f(y) \), we need to evaluate it at the endpoints of the range of \( y \):
1. **At \( y = 1 \)**:
\[ f(1) = \frac{1}{2} + \frac{1}{(1 + 2)^2} = \frac{1}{2} + \frac{1}{9} = \frac{4.5 + 1}{9} = \frac{5.5}{9} \approx 0.6111 \]

2. **At \( y = -1 \)**:
\[ f(-1) = \frac{1}{2} + \frac{1}{(-1 + 2)^2} = \frac{1}{2} + 1 = 1.5 \]

### Step 6: Determine Minimum and Maximum Values
- The minimum value occurs when \( y = 1 \), yielding \( f(1) = \frac{5}{9} \).
- The maximum value occurs when \( y = -1 \), yielding \( f(-1) = 1 \).

### Conclusion: Range of \( f(x) \)
Thus, the range of \( f(x) \) is:
\[ \left[\frac{5}{9}, 1\right] \]