How do you show that #f(x)=7x+1# and #g(x)=(x-1)/7# are inverse functions algebraically and graphically?

2 Answers
Conor M.
Apr 16, 2018

They are.

Explanation:

Using the inverse composition rule you know that if (fog)=x and (gof)=x then the functions are inverses.
#(gof)=((7x+1)-1)/7# this will result in (gof)=x
#(fog)=7(x-1)/7+1= x-1+1=x#
graphically you can check this if it seems that the functions are reflected over the line x=y.

Somebody N.
Apr 16, 2018

See below.

Explanation:

For functions:

#f(x) and g(x)#

If #f(x) and g(x)# are inverses, then:

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

#f(g(x))=7(g(x))+1=7((x-1)/7)+1=(7x-7+7)/7=x#

#g(f(x))=(f(x)-1)/7=(7x+1-1)/7=x#

#:.#

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

This makes sense if we consider what an inverse function does.

If we put a value #x# into #f(x)# we obtain a value #y#, if we then put this value #y# through the inverse function #g(x)#, we would expect to get back the value we put into #f(x)# which was #x#.

Graphically the inverse function is always a reflection of the function in the line #y=x#:

So:

#g(x)# is a reflection of #f(x)# in the line #y=x#

GRAPH:

enter image source here

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