\( \left. \begin{array} { l } { f ( x ) = \frac { x - 2 } { 4 x - x ^ { 2 } } } \\ { 4 x - x ^ { 2 } \neq 0 } \\ { x ^ { 2 } - 4 x \neq 0 } \\ { x ( x - 4 ) \neq 0 } \\ { x \neq 0 } \\ { x \neq 4 } \\ { D = \{ X E R / X \neq 0 , X \neq 4 \} } \\ { D = ( - \infty , 0 ) \cup ( 0,4 ) \cup ( 4 , \infty ) } \end{array} \right. \)

The function you provided is \[ f(x) = \frac{x - 2}{4x - x^2} \] To analyze this function, we need to determine its domain, which is the set of all \( x \) values for which the function is defined. The function is undefined when the denominator is equal to zero. 1. **Finding the zeros of the denominator:** The denominator is \( 4x - x^2 \). We can factor this expression: \[ 4x - x^2 = x(4 - x) \] Setting the denominator equal to zero gives us: \[ x(4 - x) = 0 \] This results in two solutions: \[ x = 0 \quad \text{and} \quad x = 4 \] 2. **Excluding these values from the domain:** Since the function is undefined at \( x = 0 \) and \( x = 4 \), we exclude these points from the domain. 3. **Writing the domain:** The domain of the function can be expressed in interval notation as: \[ D = (-\infty, 0) \cup (0, 4) \cup (4, \infty) \] Thus, the final domain of the function \( f(x) \) is: \[ D = \{ x \in \mathbb{R} \mid x \neq 0, x \neq 4 \} = (-\infty, 0) \cup (0, 4) \cup (4, \infty) \] This means that the function is defined for all real numbers except for \( 0 \) and \( 4 \).