Let #f(x)=5x-4 and g(x)=x+7#. Determine # color(white)("d") a: (f@g)(x), color(white)("d")b: (g@f)(x),color(white)("d")c: (f@g)(3)color(white)(...)# ?

this is intermediate algebra and I don't understand how to solve this problem regarding functions

1 Answer
Dec 1, 2017

A lot of explanation given.

Part a: #(f@g)(x)=z=5x+31#

Part b: #(g@f)(x)=5x+3#

Part c: #(f@g)(3)=15+31#

Explanation:

#color(magenta)("Explaining what it all means")#

The letters #f and g# are a sorts of names that the writer may assign to any particular expression structure they so choose.

For example, if I use the word 'wheel' you immediately know what I am talking about it and possibly a picture of a wheel occurs in your mind.

Equally the original writer of this question intended the structure #f# when applied to 'something' to mean #5xx"something" -4#

So #f# of 'something' #color(white)("ddd")#is written as:

#f("'something'")->5xx"'something'"-4#

so #f(x)->5xx x-4 color(white)("ddd")->color(white)("ddd") 5x-4#

In the same way the different structure name #g# when applied to #x# does this to #xcolor(white)("d")->color(white)("d")x+7# so we have:

#g(x)=x+7#
#g(b)=b+7#
#g("wheel")="wheel"+7# and so on
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We can nest these (put one inside the other) and one possible way of demonstrating this is:

Set #color(red)(y=g(x)=x+7)#

Set #color(blue)(z=f(color(red)(y))=5color(red)(y)-4)#

The mathematical symbols: #(f@g)(x)# is saying take the structure of #g# applied to #x# and then apply the structure of #f# to that. In a way it is a bit like substitution as I have shown above. The form #(f@g)(x)# is worded: #f# of #g# of #x#

Ok so lets do it!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("Answering part a: "(f@g)(x))#

#g# is in the right so that is the substitution.

#color(red)(y=g(x)=x+7)#

#color(blue)(z=f(color(red)(y))=5color(red)(y)-4)#

Thus by substituting for #y# we have:

#color(blue)(z=f(color(red)(y))=5color(red)((x+7))-4)#

#z =f(y)=color(white)("d") 5x+35-4 #

#(f@g)(x)=z=5x+31 ......................Equation(1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("Answering part b: "(g@f)(x))#

#f# is on the right so that comes first and is the substitution

#color(red)(y=f(x)=5x-4)#

#color(blue)(z=g(color(red)(y))= color(red)(y)+ 7 #

#color(blue)(z=color(red)(5x-4) +7)#

#z=g(y)=5x+3#

#(g@f)(x)=5x+3 ................Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(magenta)("Answering part c: "(f@g)(3))#

Using #Equation(1)#

#(f@g)(x)=5x+31#

#(f@g)(3)=5(3)+31#

#(f@g)(3)=15+31#

#(f@g)(3)=46#