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

Correct answer: ( [0,1) )

The direct answer is option A, \([0,1)\). For every real number \(x\), we have \(x^2\ge0\). Therefore the numerator is non-negative and the denominator \(x^2+1\) is always positive, so the function cannot be negative. When \(x=0\), \(f(0)=0\), so zero is included. Also, \(x^2<x^2+1\), and division by the positive denominator gives \(\frac{x^2}{x^2+1}<1\). Thus the value is always at least 0 but always less than 1. The value 1 is never reached: solving \(\frac{x^2}{x^2+1}=1\) would give \(x^2=x^2+1\), which is impossible. Option A is correct. Option B, \((0,1]\), wrongly excludes the attainable value 0 and includes the unattainable value 1. Option C, \([0,1]\), wrongly includes 1. Option D, \(\mathbb R\), is impossible because the function never gives negative values or values at least 1. Memory cue: zero may be included when the numerator can be zero; check separately whether the endpoint 1 is attained.