Question

suppose \\(h(x) = 3\sqrt3{x} + 2\\).

find two functions \\(f\\) and \\(g\\) such that \\((f \circ g)(x) = h(x)\\).

neither function can be the identity function.
(there may be more than one correct answer.)

\\(f(x) = \square\\)
\\(g(x) = \square\\)

Explanation:

Identify the inner function

$$ g(x) = \sqrt[3]{x} $$

Identify the outer function

$$ f(x) = 3x + 2 $$

Verify the composition

$$ (f \circ g)(x) = f(g(x)) = f(\sqrt[3]{x}) = 3\sqrt[3]{x} + 2 = H(x) $$

Answer:

Suppose \(H(x) = 3\sqrt[3]{x} + 2\).

Find two functions \(f\) and \(g\) such that \((f \circ g)(x) = H(x)\).

Neither function can be the identity function. (There may be more than one correct answer.)

\(f(x) =\) <blank>\(3x + 2\)</blank>

\(g(x) =\) <blank>\(\sqrt[3]{x}\)</blank>