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.)

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

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.)

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

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Definition of a Function" ], "new_concepts": [ "Function Composition", "Decomposing Functions" ], "current_concepts": [ "Definition of a Function", "Function Composition", "Decomposing Functions" ] } ## 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 $$ \begin{aligned} &f(x) = x^5 eq x\ &g(x) = 6x + 3 eq x \end{aligned} $$ 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.) $f(x) =$ $x^5$ $g(x) =$ $6x + 3$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Algebra", "Function Composition" ] }

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Question

Explanation:

Identify the inner and outer functions

Verify the non-identity condition

Answer: