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]}