Question

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Answer 1

\(\bf f(x) = 4x - 32 \qquad {\color{brown}{ g(x)}} = 4x
\\ \quad \\
(f\circ g)(x)\to f(\quad {\color{brown}{ g(x)}}\quad )=4{\color{brown}{ g(x)}} - 32\implies f(\quad {\color{brown}{ g(x)}}\quad )=4({\color{brown}{ 4x}})-32\)

Answer 2

notice above... the g(x) became the argument, that is "x"
it "took over" the "x" instance

in this case you just happen to have 1 instance of "x" in f(x)
but the argument will take over all instances

for example \(\bf f(x) = 4x - 5x^2+x^3+1 \qquad {\color{brown}{ g(x)}} = 4x
\\ \quad \\
(f\circ g)(x)\to f(\quad {\color{brown}{ g(x)}}\quad )=4{\color{brown}{ g(x)}} - 5({\color{brown}{ g(x)}})^2+({\color{brown}{ g(x)}})^3+1
\\ \quad \\
f(\quad {\color{brown}{ g(x)}}\quad )=4({\color{brown}{ 4x}}) - 5({\color{brown}{ 4x}})^2+({\color{brown}{4x}})^3+1\)

Answer 3

since g(x) is = 4x
then one can say that
f( g(x) ) = f( 4x )

Answer 4

hmmm well... dunno what you could call it... "replacement" maybe =)

Answer 5

yw