Which element of domain has the image 1 in the function \( f(x)=\frac{2 \tan x}{1+\tan x} \) ?

To determine the values of \( x \) in the domain of the function \( f(x) = \frac{2 \tan x}{1 + \tan x} \) that map to the image \( 1 \), we need to solve the equation: \[ \frac{2 \tan x}{1 + \tan x} = 1 \] **Step 1: Solve for \( \tan x \)** \[ 2 \tan x = 1 + \tan x \] \[ 2 \tan x - \tan x = 1 \] \[ \tan x = 1 \] **Step 2: Find \( x \) such that \( \tan x = 1 \)** The general solution for \( \tan x = 1 \) is: \[ x = \frac{\pi}{4} + k\pi \quad \text{where } k \text{ is any integer} \] **Step 3: Determine the Domain of \( f(x) \)** The function \( f(x) \) is defined as long as the denominator \( 1 + \tan x \) is not zero and \( \tan x \) itself is defined. Therefore, the domain excludes: \[ x \neq \frac{\pi}{2} + k\pi \quad \text{and} \quad x \neq -\frac{\pi}{4} + k\pi \quad \text{for any integer } k \] **Conclusion:** The elements of the domain that map to the image \( 1 \) under the function \( f(x) \) are all real numbers of the form: \[ x = \frac{\pi}{4} + k\pi \quad \text{where } k \text{ is any integer} \] **Final Answer:** All real numbers congruent to π⁄4 modulo π. In other words, the domain elements x satisfy x = π/4 + k π for some integer k.