Let [f:(0,1)R] be a function defined [f(x) = 1 1 - e^ - x], and [g(x) = ( f( - x) - f(x) )…

Question

Let $$f:(0,1)\to\mathbb{R}$$ be a function defined $$f(x) = {1 \over {1 - {e^{ - x}}}}$$, and $$g(x) = \left( {f( - x) - f(x)} \right)$$. Consider two statements

(I) g is an increasing function in (0, 1)

(II) g is one-one in (0, 1)

Then,

Accepted Answer

$g(x)=f(-x)-f(x)$

$$
\begin{aligned}
& =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\
& =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\
& =\frac{1+e^{x}}{1-e^{x}} \\
g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\
& =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0
\end{aligned}
$$

So both statements are correct

Solutions · 1