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What is the domain of f(x)=√((6-x)/(x+2))?

Answer and explanation

Correct answer: (-2,6]

For a real-valued square-root function, the radicand must be non-negative, and its denominator must be nonzero. Thus we require (6−x)/(x+2)≥0 with x≠−2. The critical values are x=−2, where the expression is undefined, and x=6, where the numerator is zero. A sign check gives: for x<−2, the numerator is positive and denominator negative, so the quotient is negative; for −2<x<6, both are positive, so the quotient is positive; for x>6, the quotient is negative. At x=6 the radicand is zero and is allowed. Therefore the domain is (−2,6], so option A is correct. D wrongly includes −2, while B and C select intervals where the radicand is negative.

Related tags

Real-Valued-FunctionsDomainRadical-Rational-ExpressionReal Valued FunctionsTheir Domain And RangeReal Valued Functions Their Domain And RangeRelations And FunctionsMathematicsClass 11 Mcq

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