What is the range of the function \(f(x)=x^2+1\) when \(x\in\mathbb{R}\)?

Correct answer: \([1,\infty)\)

For every real value of \(x\), \(x^2\ge 0\). Hence, \(f(x)=x^2+1\ge 1\). At \(x=0\), \(f(0)=1\), so 1 is included in the range, giving \([1,\infty)\). Option B ignores the effect of the added 1, while option D incorrectly excludes the attainable endpoint 1. Exam tip: For a quadratic of the form \(x^2+c\), find the minimum of \(x^2\) first and then add \(c\).