Find the smallest and the largest values of `tan^(-1) ((1 - x)/(1 + x)), 0 le x le 1`

Explanation

To find the smallest and largest values of the function \( f(x) = \tan^{-1} \left( \frac{1 - x}{1 + x} \right) \) for \( 0 \leq x \leq 1 \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \tan^{-1} \left( \frac{1 - x}{1 + x} \right) \]

### Step 2: Use a substitution
Let \( x = \tan(\theta) \). Then, we can express the function in terms of \( \theta \):
\[ f(\tan(\theta)) = \tan^{-1} \left( \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \right) \]

### Step 3: Apply the tangent subtraction identity
Using the identity \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \), we can rewrite the function:
\[ f(\tan(\theta)) = \tan^{-1} \left( \tan\left(\frac{\pi}{4} - \theta\right) \right) \]
This simplifies to:
\[ f(\tan(\theta)) = \frac{\pi}{4} - \theta \]

### Step 4: Determine the range of \( \theta \)
Given that \( x \) ranges from \( 0 \) to \( 1 \), we find the corresponding values of \( \theta \):
- When \( x = 0 \), \( \theta = \tan^{-1}(0) = 0 \).
- When \( x = 1 \), \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

### Step 5: Evaluate the function at the endpoints
Now we can evaluate \( f(x) \) at the endpoints of the interval:
- For \( x = 0 \):
\[ f(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]
- For \( x = 1 \):
\[ f(1) = \frac{\pi}{4} - \frac{\pi}{4} = 0 \]

### Step 6: Conclusion
Thus, the smallest value of \( f(x) \) occurs at \( x = 1 \) and is \( 0 \), and the largest value occurs at \( x = 0 \) and is \( \frac{\pi}{4} \).

### Final Answer:
- **Smallest value:** \( 0 \)
- **Largest value:** \( \frac{\pi}{4} \)