To solve the problem of finding two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] such that [tex]\((f \circ g)(x) = H(x)\)[/tex], where [tex]\( H(x) = 3 \sqrt[3]{x} - 2 \)[/tex], we need to express [tex]\( H(x) \)[/tex] as a composition of two functions.
First, recall that [tex]\((f \circ g)(x)\)[/tex] means [tex]\( f(g(x)) \)[/tex]. So, we need to find [tex]\( g(x) \)[/tex] and [tex]\( f(y) \)[/tex] (where [tex]\( y = g(x) \)[/tex]) such that:
[tex]\[ H(x) = f(g(x)) \][/tex]
Let's denote [tex]\( g(x) \)[/tex] as the inner function and [tex]\( f(y) \)[/tex] as the outer function.
1. Choose [tex]\( g(x) \)[/tex]:
Identify a function [tex]\( g(x) \)[/tex] that transforms [tex]\( x \)[/tex] into a form suitable for [tex]\( f(y) \)[/tex]. Notice that our function [tex]\( H(x) \)[/tex] involves a cube root, so a natural choice for [tex]\( g(x) \)[/tex] can be:
[tex]\[
g(x) = \sqrt[3]{x} = x^{1/3}
\][/tex]
2. Define [tex]\( f(y) \)[/tex]:
After applying [tex]\( g(x) \)[/tex], [tex]\( y = g(x) = x^{1/3} \)[/tex]. Now, we need [tex]\( f(y) \)[/tex] to transform [tex]\( y \)[/tex] into [tex]\( H(x) \)[/tex]. Thus, we seek [tex]\( f(y) \)[/tex] such that:
[tex]\[
f(g(x)) = f(x^{1/3}) = 3 \sqrt[3]{x} - 2
\][/tex]
Since [tex]\( y = x^{1/3} \)[/tex], we get:
[tex]\[
3y - 2 = 3x^{1/3} - 2
\][/tex]
Hence, the function [tex]\( f \)[/tex] that fits this transformation is:
[tex]\[
f(y) = 3y - 2
\][/tex]
So, the two functions are:
[tex]\[
\begin{array}{l}
f(y) = 3y - 2, \\
g(x) = x^{1/3}.
\end{array}
\][/tex]
Finally, verify the composition [tex]\( (f \circ g)(x) \)[/tex]:
[tex]\[
(f \circ g)(x) = f(g(x)) = f(x^{1/3}) = 3(x^{1/3}) - 2 = 3 \sqrt[3]{x} - 2.
\][/tex]
This confirms that:
[tex]\[
(f \circ g)(x) = H(x).
\][/tex]
Thus, the functions are:
[tex]\[
\begin{array}{l}
f(x) = 3x - 2, \\
g(x) = x^{1/3}.
\end{array}
\][/tex]
