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> Let $\displaystyle f = \{(x , \left( \frac{x^2}{1 + x^2} \right)) : x\in \Bbb R\}$ be a function. Determine the range of $f$.
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I did it like this:  

**Domain**: Since denominator can't be equal to zero, $1 + x^2 \neq 0 \Longrightarrow x^2 \neq -1$.

**Range**: Let $y= f(x)$

$\displaystyle y = \frac{x^2}{1 + x^2}\Longrightarrow x = \sqrt{\frac{y}{1-y}}$

$1-y \neq 0 \Longrightarrow y\neq 1$

But the correct answer given is **range** is any **positive real number x such that $0 \le x < 1$**