Let f : R to R be defined by f(x) = frac{x}{1 | vedclass

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(A) Let $y = \frac{x}{x^2 + 1}$.Rearranging the equation,we get $y(x^2 + 1) = x$,which implies $yx^2 - x + y = 0$.Since $x$ is a real number,the discr

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$[ - \frac{1}{2}, \frac{1}{2} ]$

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(A) Let $y = \frac{x}{x^2 + 1}$.Rearranging the equation,we get $y(x^2 + 1) = x$,which implies $yx^2 - x + y = 0$.Since $x$ is a real number,the discriminant $D$ of this quadratic equation in $x$ must be greater than or equal to $0$.$D = (-1)^2 - 4(y)(y) \ge 0$.$1 - 4y^2 \ge 0$.$4y^2 \le 1$,which means $y^2 \le \frac{1}{4}$.Taking the square root,we get $|y| \le \frac{1}{2}$,which implies $y \in [ - \frac{1}{2}, \frac{1}{2} ]$.Thus,the range of $f$ is $[ - \frac{1}{2}, \frac{1}{2} ]$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Let $f : R \to R$ be defined by $f(x) = \frac{x}{1 + x^2}, x \in R$. Then the range of $f$ is

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