The greatest value of the fucntion `f(x)=sin^(-1)x^2` in interval `[-1//sqrt(2),1//sqrt(2)]` is

Options

A. `pi/3`
B. `pi/2`
C. `-pi/2`
D. `pi/6`

Explanation

To find the greatest value of the function \( f(x) = \sin^{-1}(x^2) \) in the interval \([-1/\sqrt{2}, 1/\sqrt{2}]\), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(\sin^{-1}(x^2)) \]
Using the chain rule, we have:
\[ f'(x) = \frac{1}{\sqrt{1 - (x^2)^2}} \cdot \frac{d}{dx}(x^2) = \frac{1}{\sqrt{1 - x^4}} \cdot 2x = \frac{2x}{\sqrt{1 - x^4}} \]

### Step 2: Set the derivative to zero to find critical points
To find critical points, we set the derivative equal to zero:
\[ \frac{2x}{\sqrt{1 - x^4}} = 0 \]
This implies:
\[ 2x = 0 \implies x = 0 \]

### Step 3: Evaluate the function at critical points and endpoints
Now, we evaluate \( f(x) \) at the critical point \( x = 0 \) and the endpoints of the interval \( x = -\frac{1}{\sqrt{2}} \) and \( x = \frac{1}{\sqrt{2}} \).

1. **At \( x = 0 \)**:
\[ f(0) = \sin^{-1}(0^2) = \sin^{-1}(0) = 0 \]

2. **At \( x = -\frac{1}{\sqrt{2}} \)**:
\[ f\left(-\frac{1}{\sqrt{2}}\right) = \sin^{-1}\left(\left(-\frac{1}{\sqrt{2}}\right)^2\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

3. **At \( x = \frac{1}{\sqrt{2}} \)**:
\[ f\left(\frac{1}{\sqrt{2}}\right) = \sin^{-1}\left(\left(\frac{1}{\sqrt{2}}\right)^2\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

### Step 4: Compare the values
Now we compare the values obtained:
- \( f(0) = 0 \)
- \( f\left(-\frac{1}{\sqrt{2}}\right) = \frac{\pi}{6} \)
- \( f\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{6} \)

The greatest value of \( f(x) \) in the interval \([-1/\sqrt{2}, 1/\sqrt{2}]\) is:
\[ \frac{\pi}{6} \]

### Final Answer
The greatest value of the function \( f(x) = \sin^{-1}(x^2) \) in the interval \([-1/\sqrt{2}, 1/\sqrt{2}]\) is \( \frac{\pi}{6} \).

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