Find all numbers in the range of
\[f(x) = \arctan x + \arctan \frac{1 - x}{1 + x},\]expressed in radians.  Enter all the numbers, separated by commas.
From the addition formula for tangent,
\[\tan f(x) = \tan \left( \arctan x + \arctan \frac{1 - x}{1 + x} \right) = \frac{x + \frac{1 - x}{1 + x}}{1 - x \cdot \frac{1 - x}{1 + x}} = 1.\]If $x < -1,$ then $-\frac{\pi}{2} < \arctan x < -\frac{\pi}{4}.$  Also,
\[1 + \frac{1 - x}{1 + x} = \frac{2}{1 + x} < 0,\]so $\frac{1 - x}{1 + x} < -1,$ which means $-\frac{\pi}{2} < \arctan \frac{1 - x}{1 + x} < -\frac{\pi}{4}.$  Therefore, $-\pi < f(x) < -\frac{\pi}{2}.$  Since $\tan f(x) = 1,$ $f(x) = -\frac{3 \pi}{4}.$

If $x > -1,$ then $-\frac{\pi}{4} < \arctan x < \frac{\pi}{2}.$  Also,
\[1 + \frac{1 - x}{1 + x} = \frac{2}{1 + x} > 0,\]so $\frac{1 - x}{1 + x} > -1,$ which means $-\frac{\pi}{4} < \arctan \frac{1 - x}{1 + x} < \frac{\pi}{2}.$  Therefore, $-\frac{\pi}{2} < f(x) < \pi.$  Since $\tan f(x) = 1,$ $f(x) = \frac{\pi}{4}.$

Therefore, the range of $f(x)$ consists of the numbers $\boxed{-\frac{3 \pi}{4}, \frac{\pi}{4}}.$
-\frac{3\pi}{4},\frac{\pi}{4}
Level 4
Precalculus
Problem: Find all numbers in the range of
\[f(x) = \arctan x + \arctan \frac{1 - x}{1 + x},\]expressed in radians.  Enter all the numbers, separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the values of \( f(x) = \arctan x + \arctan \frac{1 - x}{1 + x} \), we need to simplify the expression. We can use the identity for the sum of arctangents:

\[
\arctan a + \arctan b = \arctan \left( \frac{a + b}{1 - ab} \right)
\]

provided that \( ab < 1 \). Here, let \( a = x \) and \( b = \frac{1 - x}{1 + x} \). We first compute the product \( ab \):

\[
ab = x \cdot \frac{1 - x}{1 + x} = \frac{x(1 - x)}{1 + x}
\]

Next, we need to check if \( ab < 1 \). The expression \( x(1 - x) \) is non-negative for \( x \in [0, 1] \) and negative for \( x < 0 \) or \( x > 1 \). Thus, \( ab < 1 \) holds for \( x \) in the range of interest.

Now, we compute \( a + b \):

\[
a + b = x + \frac{1 - x}{1 + x} = \frac{x(1 + x) + 1 - x}{1 + x} = \frac{x + x^2 + 1 - x}{1 + x} = \frac{x^2 + 1}{1 + x}
\]

Now we can substitute back into the arctangent identity:

\[
f(x) = \arctan \left( \frac{x^2 + 1}{1 + x} \right)
\]

Next, we need to analyze the behavior of \( f(x) \) as \( x \) varies. We can evaluate \( f(x) \) at specific points to find the range. 

1. For \( x = 0 \):
   \[
   f(0) = \arctan \left( \frac{0^2 + 1}{1 + 0} \right) = \arctan(1) = \frac{\pi}{4}
   \]

2. For \( x = 1 \):
   \[
   f(1) = \arctan \left( \frac{1^2 + 1}{1 + 1} \right) = \arctan(1) = \frac{\pi}{4}
   \]

3. For \( x = -1 \):
   \[
   f(-1) = \arctan \left( \frac{(-1)^2 + 1}{1 - 1} \right) \text{ is undefined, but we can analyze limits.}
   \]

4. As \( x \to -\infty \):
   \[
   f(x) \to \arctan(-\infty) + \arctan(1) = -\frac{\pi}{2} + \frac{\pi}{4} = -\frac{\pi}{4}
   \]

5. As \( x \to \infty \):
   \[
   f(x) \to \arctan(\infty) + \arctan(0) = \frac{\pi}{2} + 0 = \frac{\pi}{2}
   \]

Now, we need to find the minimum and maximum values of \( f(x) \). The function is continuous and differentiable, and we can find critical points by setting the derivative \( f'(x) \) to zero. However, we can also observe the behavior at the endpoints and specific values.

From the evaluations, we see that \( f(x) \) takes values from \( -\frac{3\pi}{4} \) to \( \frac{\pi}{4} \). The minimum occurs at \( x = -1 \) and the maximum at \( x = 0 \) or \( x = 1 \).

Thus, the final answer is \( -\frac{3\pi}{4}, \frac{\pi}{4} \).<|END_OF_LATENT|> The final answer is $-\frac{3\pi}{4},\frac{\pi}{4}$.