Range of the function f(x) = frac{x^2}{x^2 + | vedclass

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(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than t

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$[0, 1)$

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(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than the numerator.As $x 	o \pm \infty$,$y 	o 1$,but $y$ never reaches $1$ because $x^2 At $x = 0$,$y = \frac{0}{0+1} = 0$.Since $x^2 \ge 0$ and $x^2 + 1 > 0$,the value of $y$ is always non-negative.Thus,the range of the function is $[0, 1)$.

Range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

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