Find the range of the given function. \[ f(x)=1+\sqrt{x-2} \] Answer Attempt 1 out of 2 Answer Type: Interval

To determine the range of the function \( f(x)=1+\sqrt{x-2} \), we first note that the expression under the square root, \( x-2 \), must be non-negative. This gives us the condition \( x \geq 2 \). As \( x \) approaches 2, \( f(x) \) reaches its minimum value of \( 1+\sqrt{0} = 1 \). As \( x \) increases indefinitely, \( f(x) \) will also increase indefinitely. Therefore, the range of the function is \( [1, \infty) \). In real-world scenarios, this function might represent a situation where a certain quantity (like revenue or height) only starts being relevant or valid from a certain baseline value (in this case, when \( x \) is 2). For instance, if \( x \) represents temperature in Celsius starting from 2 degrees, then \( f(x) \) could represent a comfort index starting from a minimum level. Don’t forget, when dealing with square roots, it's easy to miss the domain constraints! Always check the content under the radical to determine where the function is defined. A common mistake is to assume the function exists for all \( x \) without realizing the square root's restrictions. Always set that inside part \( \geq 0 \), and solve accordingly!