Question #bd4c8
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"How do you calculate the Gibbs Free Energy change of a reaction?"
First, we need to know that #(g @ h)(x)=g[h(x)]#
In this question, we have to sub #h(x)# into it, then, replace the #x# of #g(x)# by the value of #h(x)# as the following:
#(g @ h)(x)=g[h(x)]#
#=g(3x+1)#
#=(3x+1)^2+1#
#=(3x)^2+2*3x*1+1^2+1#
#=9x^2+6x+1+1#
#=9x^2+6x+2#
Here is the answer. Hope this can help you :)
An alternative notation for, #(g@h)(x)#, is #g(h(x))#; the latter notation clearly illustrates that one should substitute #h(x)# for every instance of #x# within #g(x)#.
Start with #g(x)#:
#g(x) = x^2+1#
Substitute #h(x)# for every #x# within #g(x)#
#g(h(x)) = (h(x))^2+1#
One right side of #g(x)#, substitute the right side of #h(x)# for every instance of #h(x)#:
#g(h(x)) = (3x+1)^2+1#
Technically, we are done but it is better to simplify the equation:
#g(h(x)) = 9x^2+ 6x+2#
Returning to the other notation:
#(g@h)(x)= 9x^2+ 6x+2#
