Range? \( f(x)=\sqrt{x-2}+5 \)

The range of the function \( f(x) = \sqrt{x - 2} + 5 \) starts from the minimum value that the square root can achieve. Since the square root only produces non-negative outputs, the minimum occurs when \( x = 2 \), yielding \( f(2) = \sqrt{0} + 5 = 5 \). As \( x \) increases beyond 2, \( f(x) \) will also increase without bound, resulting in a range of \( [5, \infty) \). To determine the domain first helps in visualizing the function better. For \( f(x) \) to be defined, \( x - 2 \geq 0 \) must hold, which implies \( x \geq 2 \). This means the function behaves just like a square root shifted upwards, starting from \( (2, 5) \) and climbing ever higher, giving it a clear and defined range.