How do you compute (fog) and (gof) if #f(x)= x/(x-2)#, #g(x)=3/x#?

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Answer 1 · 2016-01-02T08:16:08

#(f@g)(x)=3/(3-2x),(g@f)(x)=(3(x-2))/x#

#(f@g)(x)# is the same as #f(g(x))#. This means that you take #g(x)#, or #3/x#, plug it in for all the spots with #x# in #f(x)#.

#f(g(x))=(3/x)/(3/x-2)#

Find a common denominator in the denominator.

#f(g(x))=(3/x)/(3/x-(2x)/x)#

Simplify.

#f(g(x))=(3/x)/((3-2x)/x)#

Multiply by #x/x#.

#f(g(x))=3/(3-2x)=(f@g)(x)#

To find #(g@f)(x)#, use a similar method: plug #x/(x-2)# into the #x# in #3/x#.

#g(f(x))=3/(x/(x-2))#

Recall that division is the same as multiplying by the reciprocal.

#g(f(x))=3((x-2)/x)#

Simplify for the final answer.

#g(f(x))=(3(x-2))/x=(g@f)(x)#