How do you find the compositions given # f(x)=8x-1# and #g(x)=x/2#?

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Answer 1 · 2016-01-07T00:26:24

#g(f(x))=g@f=(8x-1)/2=4x-1/2#
#f(g(x))=f@g=8x/2-1=4x-1#

You can think:
a)
#y=f(x)=8x-1#
#z=g(y)=y/2#
#z=h(x)=g(f(x))=(8x-1)/2=4x-1/2#

b)
#y=g(x)=x/2#
#z=f(y)=8y-1#
#z=h(x)=f(g(x))=8x/2-1=4x-1#

Remember that:

#g@f!=f@g#

Answer 2 · 2016-01-07T00:28:10

Substitute the expression for #g(x)# in place of #x# in the definition of #f(x)# to find: #(f@g)(x) = 4x-1#

Similarly find: #(g@f)(x) = 4x-1/2#

#(f@g)(x) = f(g(x)) = f(x/2) = 8(x/2)-1 = 4x-1#

#(g@f)(x) = g(f(x)) = g(8x-1) = ((8x-1))/2 = 4x-1/2#