Questions: Consider the following function. q(x) = -1/(x-2)^2 + 3 Determine the domain and range of the original function. Express your answer in interval notation.

Solution Steps

To determine the domain of the function \( q(x) = -\frac{1}{(x-2)^{2}} + 3 \), we need to identify the values of \( x \) for which the function is defined. The function is undefined where the denominator is zero, i.e., \( x = 2 \). Therefore, the domain is all real numbers except \( x = 2 \).

For the range, we analyze the behavior of the function. The term \(-\frac{1}{(x-2)^{2}}\) is always negative or zero, and it approaches zero as \( x \) moves away from 2. Thus, the maximum value of the function is 3, and it decreases without bound as \( x \) approaches 2 from either side. Therefore, the range is \((-\infty, 3)\).

The function \( q(x) = -\frac{1}{(x-2)^{2}} + 3 \) is undefined when the denominator is zero. This occurs at \( x = 2 \). Therefore, the domain of the function is all real numbers except \( x = 2 \), which can be expressed in interval notation as: \[ \text{Domain} = (-\infty, 2) \cup (2, \infty) \]

To find the range, we analyze the behavior of the function. The term \( -\frac{1}{(x-2)^{2}} \) is always negative or zero, and it approaches zero as \( x \) moves away from 2. The maximum value of \( q(x) \) occurs at \( x \) approaching \( 2 \), where \( q(x) \) approaches \( 3 \). As \( x \) moves closer to \( 2 \) from either side, \( q(x) \) decreases without bound. Thus, the range of the function is: \[ \text{Range} = (-\infty, 3) \]

Final Answer