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 \( f(x) = \tan^{-1} \left( \frac{1 - x}{1 + x} \right) \) 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 the identity for tangent
We can use the identity:
\[ \tan^{-1} \left( \frac{1 - x}{1 + x} \right) = \frac{\pi}{4} - \tan^{-1}(x) \]
This is derived from the tangent subtraction formula.

### Step 3: Substitute the identity into the function
Now substituting the identity into our function:
\[ f(x) = \frac{\pi}{4} - \tan^{-1}(x) \]

### Step 4: Determine the values of \( f(x) \) at the endpoints
Next, we evaluate \( f(x) \) at the endpoints of the interval \( [0, 1] \).

1. **At \( x = 0 \)**:
\[ f(0) = \frac{\pi}{4} - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

2. **At \( x = 1 \)**:
\[ f(1) = \frac{\pi}{4} - \tan^{-1}(1) = \frac{\pi}{4} - \frac{\pi}{4} = 0 \]

### Step 5: Identify the maximum and minimum values
From the calculations:
- The maximum value occurs at \( x = 0 \) and is \( \frac{\pi}{4} \).
- The minimum value occurs at \( x = 1 \) and is \( 0 \).

### Conclusion
Thus, the smallest and largest values of \( f(x) \) are:
- **Smallest value**: \( 0 \) (at \( x = 1 \))
- **Largest value**: \( \frac{\pi}{4} \) (at \( x = 0 \))