How do you verify that #f(x)=(3-x)/x; g(x)=3/(x+1)# are inverses?

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Answer 1 · 2017-11-30T00:18:09

If #f(x)# and #g(x)# are inverses, then both of the following must be true:

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

Check #f(g(x))#:

Start with #f(x)#:

#f(x)=(3-x)/x#

Substitute #g(x)# for for every x in #f(x)#:

#f(g(x))=(3-g(x))/g(x)#

Substitute the right of #g(x)=3/(x+1)# into every #g(x)# on the right side of the above equation:

#f(g(x))=(3-3/(x+1))/(3/(x+1))#

Multiply the right side by 1 in form of #(x+1)/(x+1)#:

#f(g(x))=(x+1)/(x+1)(3-3/(x+1))/(3/(x+1))#

This makes the embedded denominators disappear:

#f(g(x))=(3(x+1)-3)/3#

There is a common factor of #3/3# that becomes 1:

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

#f(g(x))=x#

Check #g(f(x))#

Start with #g(x)#:

#g(x)=3/(x+1)#

Substitute #f(x)# for every x:

#g(f(x))=3/(f(x)+1)#

Substitute the right side of #f(x)=(3-x)/x# for every #f(x)# on the right side of the above equation:

#g(f(x))=3/((3-x)/x+1)#

Multiply by 1 in the form of #x/x#:

#g(f(x))=x/x 3/((3-x)/x+1)#

This makes the embedded denominator disappear:

#g(f(x))= (3x)/(3-x+x)#

#-x + x# becomes 0:

#g(f(x))= (3x)/3#

#g(f(x))= x#

Both are true, therefore, they are inverses.