Let f: (0, 1) → ℝ be a function defined by f(x) = 1 1 - e ^ - x ,...

Given,

$f: \left(0, 1\right) \to \mathbb{R} be a function defined by$f\left(x\right) = \dfrac{1}{1 - {e}^{- x}}$, and$g\left(x\right) = \left(f\left(- x\right) - f\left(x\right)\right)$</p><p>Now using$g\left(x\right) = f\left(- x\right) - f\left(x\right)$we get,</p><p>$g ( x ) = \dfrac{1}{1 - {e}^{x}} - \dfrac{{e}^{x}}{{e}^{x} - 1}$</p><p>$⇒ g\left(x\right) = \dfrac{1 + {e}^{x}}{1 - {e}^{x}}$</p><p>Now differentiating the above function with respect to$x$we get,</p><p>$⇒ {g}^{'}\left(x\right) = \dfrac{2{e}^{x}}{{\left(1 - {e}^{x}\right)}^{2}} > 0

Since ${g}^{'}\left(x\right)$ is always positive

So,$g\left(x\right)$ is one-one in \left(0, 1\right)$</p><p>And$g\left(x\right)$is increasing in \left(0, 1\right)$