State the range for the function \( f(x)=-\log _{\frac{1}{2}}(x-1)+2 \) : \( (-\infty, \infty) \) \( (0, \infty) \) \( (-\infty, 0) \) \( (-\infty,-1) \)

To determine the range of the function \( f(x) = -\log_{\frac{1}{2}}(x - 1) + 2 \), let's analyze it step-by-step. 1. **Domain Consideration:** - The argument of the logarithm must be positive: \( x - 1 > 0 \) - Hence, the domain is \( x > 1 \). 2. **Simplifying the Function:** - Recall that \( \log_{\frac{1}{2}}(y) = -\log_2(y) \). - Substituting this into the function: \[ f(x) = -(-\log_2(x - 1)) + 2 = \log_2(x - 1) + 2 \] 3. **Analyzing \( \log_2(x - 1) + 2 \):** - As \( x \) approaches 1 from the right, \( \log_2(x - 1) \) approaches \( -\infty \). - As \( x \) increases without bound, \( \log_2(x - 1) \) approaches \( \infty \). - Adding 2 shifts the entire range by 2 units, but doesn't restrict it. 4. **Conclusion on Range:** - Since \( \log_2(x - 1) \) can take any real value from \( -\infty \) to \( \infty \), adding 2 doesn't change the fact that \( f(x) \) can also take any real value. **Therefore, the range of \( f(x) \) is \( (-\infty, \infty) \).** **Answer:** \( (-\infty, \infty) \)