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Answer 1 · 2017-10-10T05:59:34

#15x+35#

I'm assuming you mean #(g\circ f)#.

#f(x)=5x+9,\qquad\qquad g(x)=3x+8#

#(g\circ f)(x)\impliesg(f(x))#

So, just plug in #f(x)# for the #x# in #g(x)#:

#g(x)=3 color(red)(x)+8#

#(gcirc f)(x)=3( color(red)(5x+9))+8#

Now, just solve using simple algebra:

#15x+27+8#

#15x+35#

#\therefore\qquad (g\circ f)(x)=15x+35#

Answer 2 · 2017-10-10T06:36:19

#(g@f)(x)=15x+35#

#(g@f)(x)=g(f(x))#

In other words: replace #x# with #f(x)# in the function #g(x)#

#(g@f)(x)=3(f(x))+8#

So:

#(g@f)(x)=3(5x+9)+8#

#(g@f)(x)=15x+27+8#

#(g@f)(x)=15x+35#