Define fog(x) and gof(x). Also find their domain and range. `f(x) = tan x, x in (-pi/2, pi/2); g(x) = sqrt(1-x^2)`

Explanation

To solve the problem, we need to define the composite functions \( f \circ g(x) \) and \( g \circ f(x) \) and then find their domains and ranges.

### Step 1: Define the Functions
We are given:
- \( f(x) = \tan x \) for \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
- \( g(x) = \sqrt{1 - x^2} \)

### Step 2: Find \( f \circ g(x) \)
The composition \( f \circ g(x) \) means we substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(\sqrt{1 - x^2}) = \tan(\sqrt{1 - x^2}) \]

### Step 3: Find the Domain of \( f \circ g(x) \)
For \( f(g(x)) \) to be defined:
1. \( g(x) = \sqrt{1 - x^2} \) must be in the domain of \( f(x) \), which is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
2. The output of \( g(x) \) must also be real, which means \( 1 - x^2 \geq 0 \) or \( -1 \leq x \leq 1 \).

Thus, we need to find the range of \( g(x) \):
- The minimum value of \( g(x) \) is 0 (when \( x = \pm 1 \)).
- The maximum value of \( g(x) \) is 1 (when \( x = 0 \)).

Since \( g(x) \) outputs values in the range \( [0, 1] \), and \( \tan(x) \) is defined for \( x \) in \( [0, 1] \), we can conclude that:
- The domain of \( f \circ g(x) \) is \( [-1, 1] \).

### Step 4: Find the Range of \( f \circ g(x) \)
Since \( g(x) \) outputs values from 0 to 1, we evaluate \( f(g(x)) \):
- At \( g(0) = 1 \), \( f(1) = \tan(1) \).
- At \( g(1) = 0 \), \( f(0) = \tan(0) = 0 \).

Thus, the range of \( f \circ g(x) \) is \( [0, \tan(1)] \).

### Step 5: Find \( g \circ f(x) \)
Now we find \( g(f(x)) \):
\[ g(f(x)) = g(\tan x) = \sqrt{1 - (\tan x)^2} \]

### Step 6: Find the Domain of \( g \circ f(x) \)
For \( g(f(x)) \) to be defined:
1. \( \tan x \) must be in the domain of \( g(x) \), which is \( [-1, 1] \).
2. This means \( -1 \leq \tan x \leq 1 \).

The values of \( x \) that satisfy this condition are:
- \( x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

### Step 7: Find the Range of \( g \circ f(x) \)
Now we evaluate \( g(f(x)) \):
- At \( x = -\frac{\pi}{4} \), \( \tan(-\frac{\pi}{4}) = -1 \) so \( g(-1) = \sqrt{1 - (-1)^2} = 0 \).
- At \( x = \frac{\pi}{4} \), \( \tan(\frac{\pi}{4}) = 1 \) so \( g(1) = \sqrt{1 - 1^2} = 0 \).
- At \( x = 0 \), \( \tan(0) = 0 \) so \( g(0) = \sqrt{1 - 0^2} = 1 \).

Thus, the range of \( g \circ f(x) \) is \( [0, 1] \).

### Summary of Results
- \( f \circ g(x) = \tan(\sqrt{1 - x^2}) \)
- Domain: \( [-1, 1] \)
- Range: \( [0, \tan(1)] \)

- \( g \circ f(x) = \sqrt{1 - \tan^2 x} \)
- Domain: \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \)
- Range: \( [0, 1] \)