To compare nlogn and 2n , take logarithm both sides for simplicity in comparision..So we have :
log(nlogn) and log(2n) which can be written as :
(logn)*(logn) and nlog2 which can be further written as :
(logn)2 and n..So n is obviously larger than (logn)2
Hence nlogn < 2n
And nlogn < nlogn ..So the ordering of functions will be :
nlogn < nlogn < 2n
==> g(x) = f(x) = h(x)