Let f = left{ left(x, frac{x^2}{1+x^2} right) | vedclass

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(A) Let $y = f(x) = \frac{x^2}{1+x^2}$.Since $x^2 \ge 0$ for all $x \in R$,the numerator is non-negative and the denominator $1+x^2 \ge 1$.Thus,$y \ge

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$[0, 1)$

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(A) Let $y = f(x) = \frac{x^2}{1+x^2}$.Since $x^2 \ge 0$ for all $x \in R$,the numerator is non-negative and the denominator $1+x^2 \ge 1$.Thus,$y \ge 0$.Now,$y(1+x^2) = x^2 \implies y + yx^2 = x^2 \implies y = x^2(1-y) \implies x^2 = \frac{y}{1-y}$.Since $x^2 \ge 0$,we must have $\frac{y}{1-y} \ge 0$.This inequality holds when $0 \le y Therefore,the range of $f$ is $[0, 1)$.

Let $f = \left\{ \left(x, \frac{x^2}{1+x^2} \right) : x \in R \right\}$ be a function from $R$ into $R$. Determine the range of $f$.

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