{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-composition"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-composition-01","title":{"zh-Hant":"函數合成：受限合成的允許輸入、求值順序與不交換","en":"Function composition: the permitted inputs of a restricted composition, evaluation order and non-commutativity"},"summary":{"zh-Hant":"設 f:ℝ→ℝ，f(x)=x−3；g:[0,∞)→ℝ，g(u)=√u。(1) 求 g∘f 的允許輸入集合 D={x∈dom(f) ∣ f(x)∈dom(g)}，並證明 D=[3,∞)；(2) 依「先 f 後 g」計算 (g∘f)(7) 與 (g∘f)(4)，並說明 (g∘f)(1) 沒有定義；(3) 說明 f∘g 是 [0,∞) 上的總函數合成，並由 (f∘g)(4)=−1≠1=(g∘f)(4) 得到 g∘f≠f∘g；(4) 說明合成不是函數值的乘法。","en":"Let f:ℝ→ℝ, f(x)=x−3, and g:[0,∞)→ℝ, g(u)=√u. (1) Find the permitted inputs D={x∈dom(f) ∣ f(x)∈dom(g)} of g∘f and prove D=[3,∞); (2) evaluate (g∘f)(7) and (g∘f)(4) by 'f first, then g', and explain why (g∘f)(1) is undefined; (3) show that f∘g is a total composition on [0,∞), and conclude g∘f≠f∘g from (f∘g)(4)=−1≠1=(g∘f)(4); (4) show that composition is not multiplication of function values."},"mko_ids":["mko-composition"]}],"title":"Concept derivation guides"}