The range of the function f(x) = frac{x}{1 + | vedclass

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(B) Given the function $f(x) = \frac{x}{1 + |x|}$ for $x \in R$.Case $1$: If $x \ge 0$,then $|x| = x$. $f(x) = \frac{x}{1 + x} = \frac{x + 1 - 1}{1 +

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$(-1, 1)$

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(B) Given the function $f(x) = \frac{x}{1 + |x|}$ for $x \in R$.Case $1$: If $x \ge 0$,then $|x| = x$. $f(x) = \frac{x}{1 + x} = \frac{x + 1 - 1}{1 + x} = 1 - \frac{1}{1 + x}$.As $x$ increases from $0$ to $\infty$,$1 + x$ increases from $1$ to $\infty$,so $\frac{1}{1 + x}$ decreases from $1$ to $0$. Thus,$f(x)$ increases from $0$ to $1$. So,the range for $x \ge 0$ is $[0, 1)$.Case $2$: If $x $f(x) = \frac{x}{1 - x} = \frac{-(1 - x) + 1}{1 - x} = -1 + \frac{1}{1 - x}$.As $x$ decreases from $0$ to $-\infty$,$1 - x$ increases from $1$ to $\infty$,so $\frac{1}{1 - x}$ decreases from $1$ to $0$. Thus,$f(x)$ decreases from $0$ to $-1$. So,the range for $x Combining both cases,the range of the function is $(-1, 0) \cup [0, 1) = (-1, 1)$.

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

The range of the function $f(x) = \frac{x}{1 + |x|}, x \in R,$ is

Similar Questions

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

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