#(h@f@g)(x)# is known as a composite function. Here's how composite functions work:
let's say that #x=1#. Your function #g(x) = 1/3(x)# now produces a #y# output of #1/3# , since #g(1)=1/3(1)#. Within a composite function, the y-value of one function becomes the x-value of the next, like so:
#g(1)=1/3=>f(1/3)=17/3=>h(17/3)=17#
Therefore,
#(h@f@g)(1)=17#
Based on this, to find the function for #(h@f@g)(x)# (combined from right to left, by the way), simply replace #x# in #f(x)# with the function #g(x)#, and replace #x# in #h(x)# with the function of #(f@g)(x)#, to get #(h@f@g)(x)#.
This, simplified, is equal to #2x+15#, and therefore #(h@f@g)(1)=2(1)+15=17#
Hope that helps!
