What is the range of the function \(f(x)=\frac{x^2+4x+8}{x^2+4x+5}\)?

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

For every real \(x\), \(x^2+4x+5=(x+2)^2+1>0\), so the domain is \(\mathbb{R}\). Put \(t=(x+2)^2\ge0\). Then \(f(x)=\frac{t+4}{t+1}=1+\frac{3}{t+1}\). At \(t=0\), the value \(4\) is attained. As \(t\to\infty\), the function approaches \(1\), but it never equals \(1\). Hence the range is \((1,4]\). Option B incorrectly includes \(1\), while option C excludes the attained value \(4\). Exam tip: For such rational functions, complete the square and determine the allowed range of the substituted variable.