If \(f(x)=x+\frac{1}{x}\), where \(x>0\), what is the range of \(f\)?

Correct answer: \([2,\infty)\)

Since \(x>0\), \(f(x)-2=x+\frac{1}{x}-2=\frac{(x-1)^2}{x}\ge 0\). Thus \(f(x)\ge 2\), with equality at \(x=1\), so 2 is included in the range. Also, \(f(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), and every value \(y\ge2\) is attained. Therefore, the range is \([2,\infty)\). Exam tip: for \(x+\frac1x\) with \(x>0\), the minimum 2 can also be found directly using AM-GM. Option B is incorrect because the value 2 is attained at \(x=1\).