#f(g(x))# is the same as #(f@g)(x)#
So
#f(g(4))# is the same as #(f@g)(4)#
But
#(f@g)(x) = sqrt(1-3x^2)/(2x-8)#
#color(white)("XXXX")##color(white)("XXXX")#...or #sqrt((1-3x^2)/(2x-8))#
#color(white)("XXXX")##color(white)("XXXX")#it's hard to guess the range of the root symbol
#color(white)("XXXX")##color(white)("XXXX")#and doesn't make any difference in this case, since...
When #x=4#
#color(white)("XXXX")#we have an attempt to divide by zero
#color(white)("XXXX")#which is undefined.