Find the interval in which the function `f(x) =cos^(-1) ((1-x^(2))/(1+x^(2)))` is increasing or decreasing.

Answer

To find the interval in which the function \( f(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) is increasing or decreasing, we will follow these steps:

### Step 1: Rewrite the function
We start by rewriting the function using a trigonometric identity. We can set \( x = \tan(\theta) \). This gives us:

\[ f(x) = \cos^{-1} \left( \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \right) \]

Using the identity \( \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \), we can simplify the function:

\[ f(x) = \cos^{-1}(\cos(2\theta)) = 2\theta = 2\tan^{-1}(x) \]

### Step 2: Differentiate the function
Next, we need to find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2\tan^{-1}(x)) \]

Using the derivative of \( \tan^{-1}(x) \):

\[ f'(x) = 2 \cdot \frac{1}{1 + x^2} \]

### Step 3: Analyze the sign of the derivative
Now, we analyze the sign of \( f'(x) \):

\[ f'(x) = \frac{2}{1 + x^2} \]

Since \( 1 + x^2 > 0 \) for all \( x \in \mathbb{R} \), it follows that \( f'(x) > 0 \) for all \( x \).

### Step 4: Conclusion
Since the derivative \( f'(x) > 0 \) for all \( x \in \mathbb{R} \), we conclude that the function \( f(x) \) is increasing for all real numbers.

### Final Answer
The function \( f(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) is increasing on the interval \( (-\infty, \infty) \).

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