What is the maximum value of the function \(f(x)=3-\sqrt{x-1}\), where \(x\ge 1\)?

Correct answer: 3

Because the square root is defined for \(x\ge 1\), we have \(\sqrt{x-1}\ge 0\). Hence \(f(x)=3-\sqrt{x-1}\le 3\). At \(x=1\), \(\sqrt{x-1}=0\), so \(f(1)=3\); therefore, the maximum value is 3. The value 1 may occur for another input, but it is not the maximum. Exam tip: For an expression of the form \(a-\sqrt{g(x)}\), find its maximum by making the square-root term as small as possible, usually 0.