Let #F(x)=3x# and #g(y)=1/y#, how do you find each of the compositions and domain and range?

Question

Serenity Jones · Accepted Answer

undefined The composition of $F(g(y))$ is $F(g(y)) = 3 \cdot \frac{1}{y} = \frac{3}{y}$.

The composition of $g(F(x))$ is $g(F(x)) = \frac{1}{3x}$.

The domain of $F(g(y))$ is all real numbers except $y = 0$, and the range is all real numbers except $0$.

The domain of $g(F(x))$ is all real numbers except $x = 0$, and the range is all real numbers except $0$.

Elijah Allen · Answer

For: #(f@g)(x) #
#h(x) = [3x]_(x=g(x)=1/x) = 3*1/x=3/x#

For: #(g@f)(x) #
#r(x) = [1/x]_(x=f(x)=3x) = 1/(3x# Given: #f(x) = 3x and g(y) =1/y# Required: Composite functions: #a) =>(f@g)(x) and b) =>(g@f)(x)# Solution Strategy: - Step 1: Rewrite the composition #h(x) = (f og)(x) =>f(g(x))#. - Step 2: Replace each occurrence of x found in the outside function with the inside. - Step 3: Simplify the answer #--------------------# For: #(f@g)(x) # #color(crimson)(Step- 1)# #h(x) = f(g(x))=f(1/x)#; #color(crimson)(Step- 2)# Replace each occurrence of #x# in #f(x)# with #g(x) = 1/x# #h(x) = [3x]_(x=g(x)=1/x) = 3*1/x=3/x# The dummy variable is not relevant so you can do this in terms of #x or y or theta#a #color(crimson)(Step- 3)# function in simplest form no step 3 needed #--------------------# For: #(g@f)(x) # #color(fuchsia)(Step- 1)# #r(x) = g(f(x))=g(3x)=#; #color(fuchsia)(Step- 2)# Replace each occurrence of #x# in #g(x)# with #f(x) = 3x# #r(x) = [1/x]_(x=f(x)=3x) = 1/(3x# #color(fuchsia)(Step- 3)# function in simplest form no step 3 needed