Which ordered pair correctly classifies the domain and range of the real-valued function \(f(x)=\dfrac{x^2}{1+x^2}\)? In each ordered pair, the first set represents the domain and the second set represents the range.

Correct answer: \((\mathbb{R},[0,1))\)

For every real \(x\), \(1+x^2>0\), so the domain is \(\mathbb{R}\). Put \(t=x^2\ge0\); then \(f=\frac{t}{1+t}\). It can be 0 but can never equal 1, so the range is \([0,1)\). Exam tip: first check where the denominator is zero.