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If the domain of the function \(f(x)=\frac{1}{x^2-k}\) is \(\mathbb{R}\setminus\{-5,5\}\), what is the value of \(k\)?

Answer and explanation

Correct answer: 25

The denominator must be nonzero for the function to be defined. Setting the denominator equal to zero gives \(x^2-k=0\), or \(x^2=k\). Since the excluded values are \(x=-5\) and \(x=5\), we get \(k=5^2=25\). Option 5 is incorrect because it would exclude \(x=\pm\sqrt{5}\). Exam tip: For a rational function, exclude every value of \(x\) that makes the denominator zero.

Related tags

Real-Valued-FunctionsDomainRangeParameterRational-FunctionsReal Valued FunctionsTheir Domain And RangeReal Valued Functions Their Domain And RangeRelations And Functions

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