Question

suppose \\(h(x) = (6x + 3)^5\\).

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:

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<reasoning>

Identify the inner and outer functions

\[

$$\begin{aligned} &H(x) = (f \circ g)(x) = f(g(x)) = (6x + 3)^5\\ &g(x) = 6x + 3\\ &f(u) = u^5 \implies f(x) = x^5 \end{aligned}$$

\]

Verify the non-identity condition

\[

$$\begin{aligned} &f(x) = x^5 eq x\\ &g(x) = 6x + 3 eq x \end{aligned}$$

\]
</reasoning>

<answer>
Suppose \(H(x) = (6x + 3)^5\).

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>\(x^5\)</blank>
\(g(x) =\) <blank>\(6x + 3\)</blank>
</answer>

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"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Mathematics",
"Algebra",
"Function Composition"
]
}
</post_analysis>

Answer:

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<reasoning>

Identify the inner and outer functions

\[

$$\begin{aligned} &H(x) = (f \circ g)(x) = f(g(x)) = (6x + 3)^5\\ &g(x) = 6x + 3\\ &f(u) = u^5 \implies f(x) = x^5 \end{aligned}$$

\]

Verify the non-identity condition

\[

$$\begin{aligned} &f(x) = x^5 eq x\\ &g(x) = 6x + 3 eq x \end{aligned}$$

\]
</reasoning>

<answer>
Suppose \(H(x) = (6x + 3)^5\).

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>\(x^5\)</blank>
\(g(x) =\) <blank>\(6x + 3\)</blank>
</answer>

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"question_type": "Fill-in-the-blank",
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"Mathematics",
"Algebra",
"Function Composition"
]
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</post_analysis>