Consider f(x)=ex−e−xex+e−x=e2x−1e2x+1=1−2e2x+1
⇒f′(x)=4e2x(e2x+1)2>0,∀x∈R
⇒f(x) is an increasing function.
⇒ Domain : R, Range :(−1,1)
For f:R→(−1,1),
f(x)=ex−e−xex+e−x,f−x:(−1,1)→R
⇒x=ex−e−xex+e−x
ey=√1+x1−x⇒f′(x)=ln√1+x1−x.
Hence, given equation is equivalent to f(x)=f′(x).
⇔f(x)=x (as f is an increasing function)
⇒ln√1+x1−x=x⇒1+x1−x=e2x
Now, draw the graph of y=1+x1−x and y=e2x. They intersect each other at x=0.