Question #bd4c8

2 Answers
Yiu A.
Nov 2, 2017

#9x^2+6x+2#

Explanation:

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 :)

Douglas K.
Nov 2, 2017

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)#.

Explanation:

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#

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