To solve the problem of finding two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] such that the composition [tex]\((f \circ g)(x) = H(x)\)[/tex], where [tex]\( H(x) = (7x - 6)^3 \)[/tex], we need to decompose [tex]\( H(x) \)[/tex] into two parts: an inner function [tex]\( g \)[/tex] and an outer function [tex]\( f \)[/tex].
Given [tex]\( H(x) = (7x - 6)^3 \)[/tex], we can think of [tex]\( H(x) \)[/tex] as [tex]\( f(g(x)) \)[/tex] where:
- [tex]\( g(x) \)[/tex] transforms [tex]\( x \)[/tex] in some way,
- and [tex]\( f \)[/tex] takes the output of [tex]\( g(x) \)[/tex] and transforms it further.
A good strategy is to identify a natural "inner" part and an "outer" part of the function [tex]\( H(x) \)[/tex].
First, let's consider the inner transformation [tex]\( g(x) \)[/tex]. The expression within the parenthesis, [tex]\( 7x - 6 \)[/tex], seems like a natural candidate for [tex]\( g(x) \)[/tex], as it is a transformation of [tex]\( x \)[/tex].
Thus, we define:
[tex]\[ g(x) = 7x - 6 \][/tex]
Next, we need to define the outer function [tex]\( f \)[/tex] such that [tex]\( f \)[/tex] acts on the result of [tex]\( g(x) \)[/tex] and gives us [tex]\( H(x) \)[/tex]. Since [tex]\( H(x) = (7x - 6)^3 \)[/tex], we see that if [tex]\( g(x) = 7x - 6 \)[/tex], then [tex]\( f \)[/tex] should take [tex]\( g(x) \)[/tex] and raise it to the third power.
Thus, we define:
[tex]\[ f(x) = x^3 \][/tex]
Therefore, the two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are:
[tex]\[ f(x) = x^3 \][/tex]
[tex]\[ g(x) = 7x - 6 \][/tex]
To verify, let's compute [tex]\((f \circ g)(x)\)[/tex], which means [tex]\( f(g(x)) \)[/tex]:
[tex]\[ g(x) = 7x - 6 \][/tex]
[tex]\[ f(g(x)) = f(7x - 6) = (7x - 6)^3 \][/tex]
This matches the original function [tex]\( H(x) = (7x - 6)^3 \)[/tex], confirming that our choice of functions is correct.
So, the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] such that [tex]\((f \circ g)(x) = H(x)\)[/tex] are:
[tex]\[ f(x) = x^3 \][/tex]
[tex]\[ g(x) = 7x - 6 \][/tex]
