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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)

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