For real \(x\), if \(f(x)=\frac{x^2+4}{x^2+1}\), what is its range?

Correct answer: \((1,4]\)

Rewrite the function as \(f(x)=1+\frac{3}{x^2+1}\). Since \(x^2\geq 0\), we have \(x^2+1\geq 1\), so \(0<\frac{3}{x^2+1}\leq 3\). Hence \(1<f(x)\leq 4\), giving the range \((1,4]\). The value \(4\) is attained at \(x=0\), so it is included. The value \(1\) is approached only as \(|x|\to\infty\) and is never attained for real \(x\), so option B is incorrect. Exam tip: always test whether a range endpoint is attained or merely approached.