First we note that as e^(2x) > 0 the denominator of the quotient never vanishes, so that:
f(x) = (2e^x)/(1+e^(2x))
is defined and continuous for all x in RR.
By writing the function as:
f(x) = 2/((1+e^(2x))/e^x) = 2/(e^-x+e^x) = 1/cosh x
we can also see that the function is even and at the limits of the domain of definition we have:
lim_(x->-oo) 2/(e^-x+e^x) = lim_(x->oo) 2/(e^-x+e^x)= 0
So that f(x) has y=0 as an horizontal asymptote on both sides.
Evaluate now the first and second derivatives:
(df)/dx = d/dx (1/cosh) = -sinhx/cosh^2x
(d^2f)/dx^2 = -d/dx (sinhx/cosh^2x) = -(cosh^3x - 2sinh^2xcoshx)/cosh^4x = (2sinh^2x-cosh^2x)/cosh^3x = (sinh^2x-1)/cosh^3x
We have then that (df)/dx > 0 for x < 0 and (df)/dx < 0 for x > 0 while the only stationary point where (df)/dx = 0 is for x=0.
The function is therefore strictly increasing in (-oo,0), strictly decreasing in (0,+oo) and reaches a maximum for x=0 ov value f(0) = 1. As coshx >=1 we can see that this is an absolute maximum.
The second derivative is null for sinhx = +-1, that is for:
x = ln(sqrt2+-1)
These are two inflection points and the function is concave down in the interval (ln(sqrt2-1),ln(sqrt2+1)) and concave up outside this interval.
graph{ (2e^x)/(1+e^(2x)) [-3.59, 3.59, -1.794, 1.796]}
