Question

The range of the function $f(x)=\frac{e^{x}-e^{|x|}}{e^{x}+e^{|x|}}$ is

• A. $(-\infty, \infty)$
• B. $[0,1)$
• C. $(-1,0]$
• D. $(-1,1)$

Answer 1

Answer: (C)

$f(x) = \frac{e^x - e^{|x|}}{e^x + e^{|x|}} = \begin{cases} 0, & x \ge 0 \\[2ex] \frac{e^x - e^{|x|}}{e^x + e^{|x|}}, & x < 0 \end{cases}$
Clearly, $f(x)$ is identically xero if $x \ge 0 \dots$ (1)
If $x < 0$, let $y = f(x) = \frac{e^x - e^{-x}}{ e^x + e^{-x}} $ or
$e^{2x} = \frac{1 + y}{1 -y}$
$\because x < 0, e^{2x} < 1$ or $0 < e^{2x} < 1$
$\therefore 0 < \frac{1 +y}{1 -y} < 1$ or $\frac{1 +y}{1 -y} > 0$
and $\frac{1 +y}{1-y} < 1$
or $(y + 1)(y - 1) < 0$
or $y > 1$ or $-1 < y < 0$