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What is the range of the function \(f(x)=\sqrt{6x-x^2-5}\)?

Answer and explanation

Correct answer: [0,2]

The radicand can be rewritten as \(6x-x^2-5=4-(x-3)^2\). For the function to be real, \((x-3)^2\leq 4\), giving the domain \([1,5]\). The radicand varies from 0 to 4, so taking its square root gives \(f(x)\in[0,2]\). Option C is the range of the radicand, not of the function. Exam tip: for a square-root function, first find the range of the expression under the root and then apply the square root.

Related tags

Real-Valued-FunctionsRangeDomainRadical-FunctionsQuadratic-FunctionsReal Valued FunctionsTheir Domain And RangeReal Valued Functions Their Domain And RangeRelations And Functions

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