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

Correct answer: \([0,\infty)\)

For every real \(x\), \(x^2\geq 0\), so \(f(x)=2x^2\geq 0\). At \(x=0\), we get \(f(0)=0\). Also, for any \(y>0\), taking \(x=\sqrt{y/2}\) gives \(f(x)=y\). Thus the function attains all and only non-negative real values, so its range is \([0,\infty)\). Option B is the closest incorrect choice because it excludes \(0\). Exam tip: for \(ax^2\) with \(a>0\) and domain \(\mathbb{R}\), the minimum value is \(0\).