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If (f:[0,16]\to\mathbb{R}) is given by (f(x)=\sqrt{x}+\sqrt{16-x}), what is the maximum value of (f)?

Answer and explanation

Correct answer: (4\sqrt{2})

By symmetry the maximum occurs at (x=8), and the value is (2\sqrt{8}=4\sqrt{2}). For sums of square roots, check the balanced point.

Related tags

RangeMaximumSquare Root FunctionFunctions As A Special Kind Of RelationRelations And FunctionsMathematicsClass 11 Mcq

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