On the interval `[-(pi)/(4) , (pi)/(4)]` , the function
`f (x) = sqrt(1 + sin^(2) x)` has a maximum value of

Options

A. `0.78`
B. 1
C. `1.1`
D. `1.2`

Explanation

To find the maximum value of the function \( f(x) = \sqrt{1 + \sin^2 x} \) on the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\), we can follow these steps:

### Step 1: Identify the critical points
To find the maximum value of \( f(x) \), we first need to determine where \( \sin x \) is maximized in the given interval. The sine function is continuous and differentiable, and it reaches its maximum value at specific points.

**Hint:** The sine function reaches its maximum at \( \frac{\pi}{4} \) in the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\).

### Step 2: Evaluate \( \sin x \) at the endpoints and critical points
We will evaluate \( f(x) \) at the endpoints of the interval and at the point where \( \sin x \) is maximum.

1. **At \( x = -\frac{\pi}{4} \):**
\[ f\left(-\frac{\pi}{4}\right) = \sqrt{1 + \sin^2\left(-\frac{\pi}{4}\right)} = \sqrt{1 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}. \]

2. **At \( x = \frac{\pi}{4} \):**
\[ f\left(\frac{\pi}{4}\right) = \sqrt{1 + \sin^2\left(\frac{\pi}{4}\right)} = \sqrt{1 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}. \]

### Step 3: Compare the values
Now we compare the values of \( f(x) \) at the endpoints:
- \( f\left(-\frac{\pi}{4}\right) = \sqrt{\frac{3}{2}} \)
- \( f\left(\frac{\pi}{4}\right) = \sqrt{\frac{3}{2}} \)

Since both endpoints give the same value, we conclude that the maximum value of \( f(x) \) on the interval occurs at both endpoints.

### Step 4: Conclusion
Thus, the maximum value of \( f(x) \) on the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\) is:
\[ \sqrt{\frac{3}{2}} \approx 1.2247. \]

### Final Answer
The maximum value of \( f(x) \) on the interval \([- \frac{\pi}{4}, \frac{\pi}{4}]\) is \( \sqrt{\frac{3}{2}} \).

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