Question

Find the range of the function:
$\boxed{f(x)= \dfrac{(2^x - 2^{-x})}{(2^x + 2^{-x})}}$
​

Answer 1

Step-by-step explanation:

We have,

$y = f(x)= \dfrac{2^x - 2^{-x}}{2^x + 2^{-x}}$

$\implies \: y= \dfrac{2^x - 2^{-x}}{2^x + 2^{-x}}$

$\implies \: y \cdot {2}^{x} + y \cdot {2}^{ - x} = 2^x - 2^{-x}$

$\implies \: y \cdot {2}^{x} + \dfrac{y }{ {2}^{x} } = 2^x - \dfrac{1}{ 2^{x}}$

$\implies \: y \cdot {2}^{x} \cdot {2}^{x} + y = 2^x \cdot {2}^{x} - 1$

$\implies \: y \cdot {2}^{2x} + y = {2}^{2x} - 1$

$\implies \: (y - 1) {2}^{2x} + y + 1 = 0$

$\implies \: {2}^{2x} = - \dfrac{ y + 1 }{y - 1}$

Now, we know,

$\implies \: {2}^{2x} > 0$

$\implies \: - \dfrac{ y + 1 }{y - 1} > 0$

$\implies \: \dfrac{ y + 1 }{y - 1} < 0$

$\implies \: y \in( - 1,1)$