Let #f(x)=5x+4# and #g(x)=x−4/5# , find: a). #(f@g)(x)# ? b). #(g@f)(x)# ?

3 Answers
Feb 26, 2018

#(f ∘ g) (x) = 5x# #(g ∘ f) (x)=5x+16/5#

Explanation:

Finding #(f ∘ g) (x)# means finding #f(x)# when it is composed with #g(x)#, or #f(g(x))# . This means replacing all instances of #x# in #f(x)=5x+4# with #g(x)=x-4/5#:

#(f ∘ g) (x)=5(g(x))+4=5(x-4/5)+4=5x-4+4=5x#

Thus, #(f ∘ g) (x) = 5x#

Finding #(g ∘ f) (x)# means finding #g(x)# when it is composed with #f(x)#, or #g(f(x)).# This means replacing all instances of #x# in #g(x)=x-4/5# with #f(x)=5x+4:#

#(g ∘ f) (x) = f(x)-4/5=5x+4-4/5=5x+20/5-4/5=5x+16/5#

Thus, #(g ∘ f) (x)=5x+16/5#

Feb 26, 2018

See explanation...

Explanation:

Alright, first lets remember what #f@g# and #g@f# mean.

#f@g# is a fancy way of saying #f(g(x))# and #g@f# is a fancy way of saying #g(f(x))#. Once we realize this, these problems aren't that difficult to solve.

So #f(x)=5x+4# and #g(x)=x-4/5#

a) #f@g#

Ok lets start with the #f(x)# function

#f(x)=5x+4#

Then, we just add the #g(x)# function whenever we see an #x# in the #f(x)# function.

#f(g(x))=5g(x)+4##->##5(x-4/5)+4#

Simplify:

#f(g(x)))=(5x-4)+4# #-># #5xcancel(-4)cancel(+4)#

So therefore, #f@g=5x#

b) #g@f#

Alright, it's the same process here just it's the opposite. Let's start with the #g(x)# function.

#g(x)=x-4/5#

Then, we just add the #f(x)# function whenever we see an #x# in the #g(x)# function.

#g(f(x))=f(x)-4/5##->##(5x+4)-4/5#

Simplify:

#g(f(x))=5x+16/5#

Therefore, #g@f=5x+16/5#

Hope this helped!
~Chandler Dowd

Mar 22, 2018

For #g(x)=x-4/5# it is solved by Chandler Dowd and VNVDVI
For #g(x)=(x-4)/5#, [ requested by Widi K. ]the solution is

#color(red)( (fog)(x)=x and (gof)(x)=x)#

Explanation:

We have,#f(x)=color(red)(5x+4 ...to(1)#
#and g(x)=color(blue)((x-4)/5.......to(2)#.
Hence,
#(fog)(x)=f(g(x))#
#(fog)(x)=f(color(blue)((x-4)/5))....to#from(2)
#(fog)(x)=f(m)#,......[ take #m=(x-4)/5# ]
#(fog)(x)=color(red)(5m+4#......[Apply (1) for #x tom#]
#(fog)(x)=cancel5(color(blue)((x-4)/cancel5))+4#...[ put #m=(x-4)/5# ]
#(fog)(x)=x-4+4#
#(fog)(x)=x#

#(gof)(x)=g(f(x))#
#(gof)(x)=g(color(red)(5x+4))......to#from(1)
#(gof)(x)=g(n)........#[ take #n=5x+4#
#(gof)(x)=(color(blue)((n-4)/5))#......[ Apply (2) for #x ton#
#(gof)(x)=(5x+4-4)/5....#[ put #n=5x+4 ]#
#(gof)(x)=(5x)/5#
#(gof)(x)=x#