Question

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

Answer 1

Since $x\le |x|\Rightarrow e^x\le e^{|x|}\forall x$ [Since $e^x$ is an increasing function.]

So $e^x-e^{|x|}\le 0$ and $e^x+e^{|x|}>0$.

So $(e^x-e^{|x|})\in (-\infty ,0]$

Since $e^x-e^{|x|}>0$, $\frac{e^x-e^{|x|}}{e^x+e^{|x|}}\in (-\infty ,0]$

So range of $f\left(x\right)$ is $(-\infty ,0]$.