{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-rate-of-change-01",
  "mko_id": "mko-rate-of-change",
  "title": {
    "zh-Hant": "兩點間的平均變化率：資格、端點對調、仿射與二次公式",
    "en": "Average rate of change between distinct inputs: eligibility, swapping endpoints, the affine and quadratic formulas"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-rate-of-change-01",
    "source": {
      "mko_id": "mko-rate-of-change",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "8c12a896998e745bbd3743c80ae005d1ffbc28d69ea33072b8e08d03c3837416",
      "english_sha256": "da425c268629ac1b5344a831974cfbf2e90a380deb76eb9cf2eab84d01f97ec2"
    },
    "mode": "theorem_proof",
    "audience_prerequisites": [
      "frac_rules",
      "distributive",
      "mul_inverse_nonzero",
      "neg_mul",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "（一）已知某指定函數 f 的 f(2)=7、f(−4)=3：判定資格並求平均變化率，化為最簡正分母分數，並說明同時對調兩端點不改變結果。（二）證明 f(x)=mx+b 的平均變化率為 m，並以 ℓ(x)=−3x+7 在 −2 與 4 之間核對。（三）證明 q(x)=a(x−h)²+k（a≠0）的平均變化率為 a(x₀+x₁−2h)，以 q(x)=−2(x−3)²+5 在 1 與 4 之間核對；並以對稱端點、改變第二端點與相同輸入三個情形說明它不是導數，也不能對相同輸入填 0。",
      "en": "(1) For a specified function f with f(2)=7 and f(−4)=3: check eligibility, compute the average rate of change as a reduced fraction with positive denominator, and show that swapping both endpoints does not change it. (2) Prove that f(x)=mx+b has average rate m, and check it with ℓ(x)=−3x+7 between −2 and 4. (3) Prove that q(x)=a(x−h)²+k (a≠0) has average rate a(x₀+x₁−2h), check it with q(x)=−2(x−3)²+5 between 1 and 4, and use symmetric endpoints, a changed second endpoint and equal inputs to show that it is not a derivative and that 0 must not be assigned when the inputs coincide."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "定義（MKO）：對同一指定函數 f 的兩個合法輸入 x₀≠x₁，平均變化率 R=(f(x₁)−f(x₀))/(x₁−x₀)。x₀=x₁ 不符合資格，不以 0 或 0/0 代替。代表分式 (b−a)/(d−c) 中 a、b 是輸出 y₀、y₁，c、d 是輸入 x₀、x₁。",
          "en": "Definition (MKO): for two permitted inputs x₀≠x₁ of the same specified function f, the average rate is R=(f(x₁)−f(x₀))/(x₁−x₀). x₀=x₁ is ineligible and is not replaced by 0 or 0/0. In the representative fraction (b−a)/(d−c), a, b are the outputs y₀, y₁ and c, d the inputs x₀, x₁."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "直接兩點題：指定函數 f 只給出 f(2)=7、f(−4)=3；取 x₀=2、x₁=−4。",
          "en": "Direct two-point exercise: the specified function f is given only through f(2)=7 and f(−4)=3; take x₀=2, x₁=−4."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "ℓ:ℝ→ℝ，ℓ(x)=−3x+7（m=−3、b=7）。",
          "en": "ℓ:ℝ→ℝ, ℓ(x)=−3x+7 (m=−3, b=7)."
        }
      },
      {
        "id": "G4",
        "statement": {
          "zh-Hant": "q:ℝ→ℝ，q(x)=−2(x−3)²+5（a=−2≠0、h=3、k=5）。",
          "en": "q:ℝ→ℝ, q(x)=−2(x−3)²+5 (a=−2≠0, h=3, k=5)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "mko_rate"
        ],
        "before_tex": "f(x_0)=y_0,\\ f(x_1)=y_1",
        "after_tex": "R=\\frac{y_1-y_0}{x_1-x_0}\\ \\ \\text{only if}\\ x_0\\neq x_1",
        "relation": "definition",
        "rule": "mko_rate",
        "conditions": [
          "x_0\\neq x_1"
        ],
        "explanation": {
          "zh-Hant": "依 MKO 定義平均變化率：輸出差除以輸入差。資格條件有兩個：兩端點是同一指定函數的合法輸入，且 x₀≠x₁，使分母不為零。代表分式 (b−a)/(d−c) 的 a、b 是輸出，c、d 是輸入，不與頂點式的 a、h、k 混用。",
          "en": "Define the average rate as in the MKO: the output difference divided by the input difference. There are two eligibility conditions: both endpoints are permitted inputs of the same specified function, and x₀≠x₁, so the denominator is nonzero. In the representative fraction (b−a)/(d−c), a, b are outputs and c, d inputs, not to be confused with the vertex-form a, h, k."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "mko_rate"
          ],
          "explanation": {
            "zh-Hant": "只是定義與資格條件；x₀=x₁ 時不定義任何值。",
            "en": "This is only the definition and its eligibility condition; for x₀=x₁ no value is defined."
          }
        },
        "leaf_basis": "mko_rate + def_intro"
      },
      {
        "id": "S02",
        "uses": [
          "S01",
          "G2"
        ],
        "before_tex": "x_0=2,\\ x_1=-4",
        "after_tex": "x_1-x_0=-6\\neq0",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩端點都是同一個 f 的輸入，且 −4−2=−6≠0，符合資格。分母為負並不違反資格，只需在化簡時把它轉為正分母。",
          "en": "Both endpoints are inputs of the same f and −4−2=−6≠0, so the pair is eligible. A negative denominator does not violate eligibility; it only has to be made positive when simplifying."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算輸入差並與 0 比較。",
            "en": "Exact computation of the input difference and comparison with 0."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "G2"
        ],
        "before_tex": "R=\\frac{3-7}{-4-2}",
        "after_tex": "\\frac{2}{3}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "-6\\neq0"
        ],
        "explanation": {
          "zh-Hant": "只用明示的輸出值 f(2)=7、f(−4)=3（MKO a1）：R=(3−7)/(−4−2)。子步 S03.1、S03.2 先算出 (−4)/(−6)，再化為最簡正分母分數 2/3。",
          "en": "Using only the given outputs f(2)=7 and f(−4)=3 (MKO a1): R=(3−7)/(−4−2). Sub-steps S03.1 and S03.2 first get (−4)/(−6) and then reduce it to the fraction 2/3 with positive denominator."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "每一步都是同一個有理數的不同寫法，雙向成立。",
            "en": "Each step rewrites the same rational number, so it holds both ways."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S03.1",
        "uses": [
          "S02"
        ],
        "before_tex": "\\frac{3-7}{-4-2}",
        "after_tex": "\\frac{-4}{-6}",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分子 3−7=−4，分母 −4−2=−6。",
          "en": "Numerator 3−7=−4, denominator −4−2=−6."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "整數減法，可逆向核對。",
            "en": "Integer subtraction, checkable backwards."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S03.2",
        "uses": [
          "S03.1",
          "frac_rules"
        ],
        "before_tex": "\\frac{-4}{-6}",
        "after_tex": "\\frac{4}{6}=\\frac{2}{3}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "分子分母同乘 −1（非零）得 4/6，分母轉正；再同除 gcd(4,6)=2 得 2/3。gcd(2,3)=1，分母 3>0，是最簡正分母形式。",
          "en": "Multiply numerator and denominator by −1 (nonzero) to get 4/6 with a positive denominator; then divide both by gcd(4,6)=2 to get 2/3. gcd(2,3)=1 and the denominator 3>0, so this is the reduced form with positive denominator."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "a/b=(ak)/(bk)（k≠0）雙向成立。",
            "en": "a/b=(ak)/(bk) for k≠0 holds both ways."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03"
        ],
        "before_tex": "\\frac{y_0-y_1}{x_0-x_1}=\\frac{7-3}{2-(-4)}",
        "after_tex": "\\frac{4}{6}=\\frac{2}{3}",
        "relation": "equivalent",
        "rule": "L_swap",
        "conditions": [
          "x_0\\neq x_1"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_swap，同時對調兩個端點（分子與分母一起換號），商不變：(7−3)/(2−(−4))=4/6=2/3，與 S03 相同。證明在 S04.1。",
          "en": "By Lemma L_swap, swapping both endpoints together (numerator and denominator both change sign) leaves the quotient unchanged: (7−3)/(2−(−4))=4/6=2/3, the same as S03. The proof is in S04.1."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_swap"
          ],
          "explanation": {
            "zh-Hant": "引理是等式，雙向成立。",
            "en": "The lemma is an equation and holds both ways."
          }
        },
        "leaf_basis": "frac_rules + neg_mul"
      },
      {
        "id": "S04.1",
        "uses": [
          "frac_rules",
          "neg_mul"
        ],
        "before_tex": "\\frac{y_0-y_1}{x_0-x_1}",
        "after_tex": "\\frac{-(y_1-y_0)}{-(x_1-x_0)}=\\frac{y_1-y_0}{x_1-x_0}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "x_1-x_0\\neq0"
        ],
        "explanation": {
          "zh-Hant": "y₀−y₁=−(y₁−y₀)，x₀−x₁=−(x₁−x₀)；再用 a/b=(a·(−1))/(b·(−1))（−1≠0、b≠0）。此步證得引理 L_swap。只換其中一對會使商變號。",
          "en": "y₀−y₁=−(y₁−y₀) and x₀−x₁=−(x₁−x₀); then use a/b=(a·(−1))/(b·(−1)) (−1≠0, b≠0). This step proves Lemma L_swap. Swapping only one of the two pairs would change the sign of the quotient."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "兩條規則都是恆等式，前提只有分母非零。",
            "en": "Both rules are identities whose only condition is a nonzero denominator."
          }
        },
        "leaf_basis": "frac_rules + neg_mul"
      },
      {
        "id": "S05",
        "uses": [
          "S01"
        ],
        "before_tex": "\\frac{(mx_1+b)-(mx_0+b)}{x_1-x_0}",
        "after_tex": "m",
        "relation": "equivalent",
        "rule": "L_linear_rate",
        "conditions": [
          "x_0\\neq x_1"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_linear_rate：仿射函數在任意兩個不同輸入之間的平均變化率都是斜率 m。證明在 S05.1、S05.2；m=0 時得 0，對應常數函數。",
          "en": "By Lemma L_linear_rate, an affine function has average rate equal to its slope m between any two different inputs. The proof is in S05.1 and S05.2; m=0 gives 0, the constant-function case."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_linear_rate"
          ],
          "explanation": {
            "zh-Hant": "引理是等式，雙向成立。",
            "en": "The lemma is an equation and holds both ways."
          }
        },
        "leaf_basis": "distributive + mul_inverse_nonzero"
      },
      {
        "id": "S05.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "(mx_1+b)-(mx_0+b)",
        "after_tex": "m(x_1-x_0)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "b 互相抵消，剩 mx₁−mx₀，再以分配律提出 m。",
          "en": "The b terms cancel, leaving mx₁−mx₀, and distributivity factors out m."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity; it holds both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm"
      },
      {
        "id": "S05.2",
        "uses": [
          "S05.1",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{m(x_1-x_0)}{x_1-x_0}",
        "after_tex": "m",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "x_1-x_0\\neq0"
        ],
        "explanation": {
          "zh-Hant": "因 x₁−x₀≠0，可約去：m(x₁−x₀)·1/(x₁−x₀)=m。若 x₁=x₀，不能約，正是資格條件的作用。S05.1 與 S05.2 合起來證得引理 L_linear_rate。",
          "en": "Since x₁−x₀≠0, it cancels: m(x₁−x₀)·1/(x₁−x₀)=m. If x₁=x₀ it could not be cancelled, which is exactly what the eligibility condition is for. S05.1 and S05.2 together prove Lemma L_linear_rate."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "乘以 (x₁−x₀) 與其倒數互逆，前提 x₁−x₀≠0。",
            "en": "Multiplying by (x₁−x₀) and by its reciprocal undo each other, given x₁−x₀≠0."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + mul_assoc + mul_one"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G3"
        ],
        "before_tex": "\\ell(-2)=13,\\ \\ell(4)=-5",
        "after_tex": "\\frac{-5-13}{4-(-2)}=\\frac{-18}{6}=-3=m",
        "relation": "implies",
        "rule": "L_linear_rate",
        "conditions": [
          "4\\neq-2"
        ],
        "explanation": {
          "zh-Hant": "ℓ(−2)=−3·(−2)+7=13，ℓ(4)=−3·4+7=−5。直接計算 (−5−13)/(4−(−2))=−18/6=−3，與 S05 的 m=−3 一致。",
          "en": "ℓ(−2)=−3·(−2)+7=13 and ℓ(4)=−3·4+7=−5. Direct computation gives (−5−13)/(4−(−2))=−18/6=−3, agreeing with m=−3 from S05."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一般結論來自 S05；數值只是核對。",
            "en": "The general conclusion comes from S05; the numbers are only a check."
          }
        },
        "leaf_basis": "mko_rate + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S01"
        ],
        "before_tex": "\\frac{q(x_1)-q(x_0)}{x_1-x_0}",
        "after_tex": "a(x_0+x_1-2h)",
        "relation": "equivalent",
        "rule": "L_quad_rate",
        "conditions": [
          "x_0\\neq x_1",
          "a\\neq0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_quad_rate：二次函數的平均變化率是 a(x₀+x₁−2h)，同時依賴兩個端點。證明在 S07.1 至 S07.3：k 抵消、平方差分解、約去 x₁−x₀。",
          "en": "By Lemma L_quad_rate, the average rate of a quadratic is a(x₀+x₁−2h), depending on both endpoints. The proof is in S07.1 to S07.3: k cancels, the difference of squares factors, and x₁−x₀ cancels."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_quad_rate"
          ],
          "explanation": {
            "zh-Hant": "引理由三個等式子步證成，雙向成立。",
            "en": "The lemma is proved by three equational sub-steps and holds both ways."
          }
        },
        "leaf_basis": "distributive + diff_squares + mul_inverse_nonzero"
      },
      {
        "id": "S07.1",
        "uses": [
          "distributive"
        ],
        "before_tex": "q(x_1)-q(x_0)",
        "after_tex": "a\\bigl[(x_1-h)^2-(x_0-h)^2\\bigr]",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(a(x₁−h)²+k)−(a(x₀−h)²+k)：k 抵消（k 只平移輸出），再以分配律提出 a。",
          "en": "(a(x₁−h)²+k)−(a(x₀−h)²+k): k cancels (it only shifts the output), and distributivity factors out a."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "恆等式，雙向成立。",
            "en": "An identity; it holds both ways."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "diff_squares"
        ],
        "before_tex": "(x_1-h)^2-(x_0-h)^2",
        "after_tex": "(x_1-x_0)(x_0+x_1-2h)",
        "relation": "equivalent",
        "rule": "diff_squares",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 u=x₁−h、v=x₀−h：u−v=x₁−x₀，u+v=x₀+x₁−2h，依平方差 u²−v²=(u−v)(u+v)。",
          "en": "Take u=x₁−h, v=x₀−h: u−v=x₁−x₀ and u+v=x₀+x₁−2h, so by the difference of squares u²−v²=(u−v)(u+v)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "diff_squares",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "平方差是恆等式，雙向成立。",
            "en": "The difference of squares is an identity and holds both ways."
          }
        },
        "leaf_basis": "diff_squares + distributive + add_inverse"
      },
      {
        "id": "S07.3",
        "uses": [
          "S07.2",
          "mul_inverse_nonzero"
        ],
        "before_tex": "\\frac{a(x_1-x_0)(x_0+x_1-2h)}{x_1-x_0}",
        "after_tex": "a(x_0+x_1-2h)",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "x_1-x_0\\neq0"
        ],
        "explanation": {
          "zh-Hant": "因 x₁−x₀≠0，約去該因子。S07.1 至 S07.3 合起來證得引理 L_quad_rate。這個化簡後的式子仍是兩端點間的差商，不是某一個端點的瞬時變化率。",
          "en": "Since x₁−x₀≠0, the factor cancels. S07.1 to S07.3 together prove Lemma L_quad_rate. The simplified expression is still the difference quotient between two endpoints, not an instantaneous rate at either endpoint."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "約分可逆，前提 x₁−x₀≠0。",
            "en": "Cancellation is reversible, given x₁−x₀≠0."
          }
        },
        "leaf_basis": "mul_inverse_nonzero + mul_comm + mul_one"
      },
      {
        "id": "S08",
        "uses": [
          "S01",
          "G4"
        ],
        "before_tex": "q(1)=-3,\\ q(4)=3",
        "after_tex": "\\frac{3-(-3)}{4-1}=\\frac{6}{3}=2",
        "relation": "implies",
        "rule": "mko_rate",
        "conditions": [
          "4\\neq1"
        ],
        "explanation": {
          "zh-Hant": "q(1)=−2·(−2)²+5=−3，q(4)=−2·1²+5=3。直接計算 (3−(−3))/(4−1)=6/3=2。",
          "en": "q(1)=−2·(−2)²+5=−3 and q(4)=−2·1²+5=3. Direct computation: (3−(−3))/(4−1)=6/3=2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_rate",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "資格 4≠1 成立；精確計算。",
            "en": "Eligibility 4≠1 holds; exact arithmetic."
          }
        },
        "leaf_basis": "mko_rate + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S07",
          "S08"
        ],
        "before_tex": "a(x_0+x_1-2h)\\ \\text{at}\\ a=-2,\\ h=3,\\ x_0=1,\\ x_1=4",
        "after_tex": "-2(1+4-6)=2",
        "relation": "implies",
        "rule": "L_quad_rate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入公式：−2·(1+4−6)=−2·(−1)=2，與 S08 的直接計算一致。",
          "en": "Substitute into the formula: −2·(1+4−6)=−2·(−1)=2, agreeing with the direct computation in S08."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "公式由 S07 證明；兩種算法一致只是核對。",
            "en": "The formula is proved in S07; agreement of the two computations is only a check."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S07",
          "G4"
        ],
        "before_tex": "x_0=1,\\ x_1=5=2h-1",
        "after_tex": "R=-2(1+5-6)=0,\\ \\text{yet}\\ q(3)=5\\neq-3=q(1)",
        "relation": "implies",
        "rule": "L_quad_rate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對稱端點 1 與 5（關於 h=3 對稱）：R=−2·0=0，直接也有 q(5)−q(1)=−3−(−3)=0。但 q 不是常數，例如 q(3)=5≠−3。平均變化率為零只說明兩端點輸出相等，不說明中間的行為。",
          "en": "Symmetric endpoints 1 and 5 (mirror images about h=3): R=−2·0=0, and directly q(5)−q(1)=−3−(−3)=0. But q is not constant, for example q(3)=5≠−3. A zero average rate only says the two endpoint outputs are equal; it says nothing about what happens in between."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個反例（q(3)≠q(1)）就足以否定「平均變化率為零 ⇒ 常數」。",
            "en": "One counterexample (q(3)≠q(1)) suffices to refute 'zero average rate ⇒ constant'."
          }
        },
        "leaf_basis": "numeric_evaluation + quantifier_negation"
      },
      {
        "id": "S11",
        "uses": [
          "S07",
          "S09"
        ],
        "before_tex": "x_0=1,\\ x_1=2",
        "after_tex": "R=-2(1+2-6)=6\\neq2",
        "relation": "implies",
        "rule": "L_quad_rate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "保持 x₀=1 而把第二端點改成 2：R=−2·(−3)=6；直接 (q(2)−q(1))/(2−1)=(3−(−3))/1=6。與 x₁=4 時的 2 不同，所以平均變化率不是只屬於 x₀=1 這一點的量；它是兩端點間的有限差商，沒有取任何極限，不能叫作導數。",
          "en": "Keep x₀=1 and change the second endpoint to 2: R=−2·(−3)=6; directly (q(2)−q(1))/(2−1)=(3−(−3))/1=6. This differs from the value 2 for x₁=4, so the average rate is not a quantity belonging to the single point x₀=1; it is a finite difference quotient between two endpoints, no limit is taken, and it may not be called a derivative."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S07",
            "S09",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只需兩個不同的值就證明它依賴第二端點；本稿不計算導數本身。",
            "en": "Two different values suffice to show the dependence on the second endpoint; the guide does not compute the derivative itself."
          }
        },
        "leaf_basis": "numeric_evaluation + quantifier_negation"
      },
      {
        "id": "S12",
        "uses": [
          "S01",
          "G4"
        ],
        "before_tex": "x_0=x_1=2",
        "after_tex": "x_1-x_0=0:\\ \\text{ineligible, no value assigned}",
        "relation": "implies",
        "rule": "mko_rate",
        "conditions": [],
        "explanation": {
          "zh-Hant": "相同輸入 x₀=x₁=2：分母 2−2=0，不符合資格。分子 q(2)−q(2)=0，形式上是 0/0，不是一個數；依 MKO 不以 0 或 0/0 代替，本條目對這組端點不給任何值。",
          "en": "Equal inputs x₀=x₁=2: the denominator 2−2=0, so the pair is ineligible. The numerator q(2)−q(2)=0, so formally it is 0/0, which is not a number; by the MKO it is not replaced by 0 or 0/0, and this entry assigns no value to this pair."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_rate",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "資格條件本身就排除此情形；不宣稱任何極限或導數。",
            "en": "The eligibility condition itself excludes this case; no limit or derivative is claimed."
          }
        },
        "leaf_basis": "mko_rate + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S04",
          "S05",
          "S06",
          "S07",
          "S09",
          "S10",
          "S11",
          "S12"
        ],
        "before_tex": "\\text{S02 to S12}",
        "after_tex": "R=\\tfrac23;\\ R_{\\ell}=m=-3;\\ R_q=a(x_0+x_1-2h),\\ R_q(1,4)=2;\\ x_0=x_1\\ \\text{ineligible}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "整理：直接兩點題的平均變化率為 2/3，對調兩端點不變；仿射函數的平均變化率恆為 m；二次函數為 a(x₀+x₁−2h)，在 1 與 4 之間為 2；對稱端點可得 0 而函數非常數；改變第二端點會改變結果，所以它不是導數；相同輸入不給值。",
          "en": "Summary: the direct two-point rate is 2/3 and does not change when both endpoints are swapped; an affine function always has rate m; a quadratic has rate a(x₀+x₁−2h), equal to 2 between 1 and 4; symmetric endpoints can give 0 although the function is not constant; changing the second endpoint changes the result, so it is not a derivative; equal inputs get no value."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S03",
            "S04",
            "S05",
            "S06",
            "S07",
            "S09",
            "S10",
            "S11",
            "S12",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G 為 goal 的三部分；足夠條件 H 為各步結論的合取，皆已正向證成，H⇒G 只用合取引入。",
            "en": "Goal G is the three parts of the goal; the sufficient condition H is the conjunction of the step results, all proved forwards, and H⇒G uses only conjunction introduction."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "f(2)=7、f(−4)=3 時平均變化率為 (3−7)/(−4−2)=2/3，同時對調兩端點不變。f(x)=mx+b 在任意 x₀≠x₁ 之間的平均變化率為 m（ℓ 的例子為 −3）。q(x)=a(x−h)²+k 的平均變化率為 a(x₀+x₁−2h)；q(x)=−2(x−3)²+5 在 1 與 4 之間為 2，在對稱端點 1 與 5 之間為 0（而 q 非常數），在 1 與 2 之間為 6，所以它依賴兩端點、不是導數。x₀=x₁ 不符合資格，不給值，也不填 0。",
      "en": "With f(2)=7 and f(−4)=3 the average rate is (3−7)/(−4−2)=2/3, unchanged when both endpoints are swapped. f(x)=mx+b has average rate m between any x₀≠x₁ (−3 for ℓ). q(x)=a(x−h)²+k has average rate a(x₀+x₁−2h); for q(x)=−2(x−3)²+5 it is 2 between 1 and 4, 0 between the symmetric endpoints 1 and 5 (though q is not constant), and 6 between 1 and 2, so it depends on both endpoints and is not a derivative. x₀=x₁ is ineligible: no value is assigned; in particular, the answer is not 0."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "x₀=x₁=2 時，q 的平均變化率是 0",
          "en": "For x₀=x₁=2 the average rate of q is 0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "分母為 0，不符合資格；0/0 不是數，不能以 0 代替。",
          "en": "The denominator is 0, so the pair is ineligible; 0/0 is not a number and may not be replaced by 0."
        }
      },
      {
        "claim": {
          "zh-Hant": "(q(4)−q(1))/(1−4)=−2 是 1 與 4 之間的平均變化率",
          "en": "(q(4)−q(1))/(1−4)=−2 is the average rate between 1 and 4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "只反轉了分母；正確為 (q(4)−q(1))/(4−1)=2。",
          "en": "Only the denominator was reversed; the correct value is (q(4)−q(1))/(4−1)=2."
        }
      },
      {
        "claim": {
          "zh-Hant": "q 在 1 與 5 之間平均變化率為 0，所以 q 在 [1,5] 上是常數",
          "en": "q has average rate 0 between 1 and 5, so q is constant on [1,5]"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "q(3)=5≠−3=q(1)。",
          "en": "q(3)=5≠−3=q(1)."
        }
      },
      {
        "claim": {
          "zh-Hant": "只由 1 到 4 的平均變化率等於 2，未提供任何導數定義或極限證書，就已證明 q 在 x=1 的導數為 2",
          "en": "The average rate from 1 to 4 is 2, so this alone proves that the derivative of q at x=1 is 2, without a derivative definition or limit certificate"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "本稿只算出指定兩端點的有限商；改成 x₁=2 時得到 6。這不足以作為某點導數值的證書，因此拒絕上述推論。本稿沒有計算 q′(1)，也不以「兩種程序不同」單獨證明其數值不等；仿射函數的平均變化率與導數值甚至可能相同。",
          "en": "This guide computes a finite quotient for two specified endpoints; changing x₁ to 2 gives 6. Such a computation is not a certificate of a derivative value at one point, so the stated inference is rejected. The guide does not compute q′(1), nor does it infer numerical inequality merely from different procedures; for an affine function the two procedures can even give the same value."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "def_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。",
          "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_rate",
        "kind": "cited",
        "source": {
          "mko_id": "mko-rate-of-change",
          "sha256": "8c12a896998e745bbd3743c80ae005d1ffbc28d69ea33072b8e08d03c3837416",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "對已知兩個函數值 f(x0)=y0、f(x1)=y1 且 x0≠x1，兩點間平均變化率為 (y1−y0)/(x1−x0)。若 x0=x1，此商不具本條目資格，不以零或 0/0 代替。對 f(x)=mx+b，該商為 m；對 q(x)=a(x−h)²+k，該商為 a(x0+x1−2h)。",
          "en": "For two permitted inputs x0≠x1 of the same specified function f, with f(x0)=y0 and f(x1)=y1, the average rate of change is (y1−y0)/(x1−x0). Equal inputs fail this entry's eligibility condition; do not replace their quotient with zero or 0/0. For f(x)=mx+b on R it equals m. For q(x)=a(x−h)²+k with a≠0 it equals a(x0+x1−2h). This finite endpoint quotient is not a derivative."
        }
      },
      {
        "id": "diff_squares",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "平方差：對所有實數 u、v，u²−v²=(u−v)(u+v)（由分配律展開右邊即得）。",
          "en": "Difference of squares: for all reals u, v, u²−v²=(u−v)(u+v) (expand the right side by distributivity)."
        }
      },
      {
        "id": "L_swap",
        "kind": "lemma",
        "proved_in": [
          "S04.1"
        ],
        "statement": {
          "zh-Hant": "對 x₀≠x₁：(y₀−y₁)/(x₀−x₁)=(y₁−y₀)/(x₁−x₀)。",
          "en": "For x₀≠x₁: (y₀−y₁)/(x₀−x₁)=(y₁−y₀)/(x₁−x₀)."
        }
      },
      {
        "id": "L_linear_rate",
        "kind": "lemma",
        "proved_in": [
          "S05.1",
          "S05.2"
        ],
        "statement": {
          "zh-Hant": "對 f(x)=mx+b 與 x₀≠x₁：(f(x₁)−f(x₀))/(x₁−x₀)=m。",
          "en": "For f(x)=mx+b and x₀≠x₁: (f(x₁)−f(x₀))/(x₁−x₀)=m."
        }
      },
      {
        "id": "L_quad_rate",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2",
          "S07.3"
        ],
        "statement": {
          "zh-Hant": "對 q(x)=a(x−h)²+k 與 x₀≠x₁：(q(x₁)−q(x₀))/(x₁−x₀)=a(x₀+x₁−2h)。",
          "en": "For q(x)=a(x−h)²+k and x₀≠x₁: (q(x₁)−q(x₀))/(x₁−x₀)=a(x₀+x₁−2h)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "-4-2",
        "rhs": "-6"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "(3-7)/(-4-2)",
        "rhs": "2/3"
      },
      {
        "id": "N3",
        "step": "S03.2",
        "lhs": "gcd(2,3)",
        "rhs": "1"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "(7-3)/(2-(-4))",
        "rhs": "2/3"
      },
      {
        "id": "N5",
        "step": "S05.1",
        "vars": [
          "m",
          "b",
          "x0",
          "x1"
        ],
        "degree": 1,
        "lhs": "(m*x1+b)-(m*x0+b)",
        "rhs": "m*(x1-x0)"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "-3*(-2)+7",
        "rhs": "13"
      },
      {
        "id": "N7",
        "step": "S06",
        "lhs": "-3*4+7",
        "rhs": "-5"
      },
      {
        "id": "N8",
        "step": "S06",
        "lhs": "(-5-13)/(4-(-2))",
        "rhs": "-3"
      },
      {
        "id": "N9",
        "step": "S07",
        "vars": [
          "a",
          "h",
          "k",
          "x0",
          "x1"
        ],
        "degree": 2,
        "lhs": "(a*(x1-h)**2+k)-(a*(x0-h)**2+k)",
        "rhs": "a*(x1-x0)*(x0+x1-2*h)"
      },
      {
        "id": "N10",
        "step": "S07.2",
        "vars": [
          "u",
          "v"
        ],
        "degree": 2,
        "lhs": "u**2-v**2",
        "rhs": "(u-v)*(u+v)"
      },
      {
        "id": "N11",
        "step": "S08",
        "lhs": "-2*(1-3)**2+5",
        "rhs": "-3"
      },
      {
        "id": "N12",
        "step": "S08",
        "lhs": "-2*(4-3)**2+5",
        "rhs": "3"
      },
      {
        "id": "N13",
        "step": "S08",
        "lhs": "(3-(-3))/(4-1)",
        "rhs": "2"
      },
      {
        "id": "N14",
        "step": "S09",
        "lhs": "-2*(1+4-2*3)",
        "rhs": "2"
      },
      {
        "id": "N15",
        "step": "S10",
        "lhs": "-2*(1+5-2*3)",
        "rhs": "0"
      },
      {
        "id": "N16",
        "step": "S10",
        "lhs": "(-2*(5-3)**2+5)-(-2*(1-3)**2+5)",
        "rhs": "0"
      },
      {
        "id": "N17",
        "step": "S10",
        "lhs": "-2*(3-3)**2+5",
        "relation": "!=",
        "rhs": "-3"
      },
      {
        "id": "N18",
        "step": "S11",
        "lhs": "-2*(1+2-2*3)",
        "rhs": "6"
      },
      {
        "id": "N19",
        "step": "S11",
        "lhs": "((-2*(2-3)**2+5)-(-2*(1-3)**2+5))/(2-1)",
        "rhs": "6"
      },
      {
        "id": "N20",
        "step": "S12",
        "lhs": "2-2",
        "rhs": "0"
      },
      {
        "id": "N21",
        "step": "S04",
        "lhs": "(3-(-3))/(1-4)",
        "rhs": "-2"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 範圍內：兩端點是同一指定函數的合法輸入且 x₀≠x₁，整數端點與係數，答案化為最簡正分母分數（a1、a2），二次切片要求 a≠0（a3）。仿射與二次公式以代數恆等式對所有合格端點證明。本稿不定義、不計算導數，只說明平均變化率不是某一點的量；不改動 canonical MKO 或英文 edition。",
      "en": "Inside the MKO: both endpoints are permitted inputs of the same specified function with x₀≠x₁, integer endpoints and coefficients, answers as reduced fractions with positive denominators (a1, a2), and a≠0 in the quadratic slice (a3). The affine and quadratic formulas are proved by algebraic identities for all eligible endpoints. This guide neither defines nor computes derivatives; it only shows that the average rate is not a quantity of a single point; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "x_0,x_1",
        "兩個輸入端點，x_0\\neq x_1",
        "the two input endpoints, x_0≠x_1"
      ],
      [
        "y_0=f(x_0),\\ y_1=f(x_1)",
        "對應的輸出",
        "the corresponding outputs"
      ],
      [
        "R(f;x_0,x_1)=\\frac{y_1-y_0}{x_1-x_0}",
        "平均變化率",
        "the average rate of change"
      ],
      [
        "\\ell(x)=-3x+7",
        "一次函數例子",
        "the degree-one affine example"
      ],
      [
        "q(x)=-2(x-3)^2+5",
        "二次函數例子",
        "the quadratic example"
      ]
    ],
    "md_error_controls": [
      {
        "title": "相同輸入不能填零 / Do not assign zero when the inputs coincide",
        "zh": "MKO 的誤解之一是相同輸入也把差商填 0。x₀=x₁=2 時分母為 0，商是 0/0，不是一個數；本條目規定此時不符合資格、不給值（S12）。",
        "en": "One misconception identified in the MKO is assigning 0 when the inputs coincide. For x₀=x₁=2 the denominator is 0 and the quotient is 0/0, which is not a number; this entry declares the pair ineligible and gives no value (S12)."
      },
      {
        "title": "平均變化率不是導數 / The average rate is not a derivative",
        "zh": "另一個誤解是把平均變化率直接叫導數。從 x₀=1 出發，到 x₁=4 得 2、到 x₁=2 得 6（S11），所以它不是只屬於一點的量；對稱端點 1、5 得 0 而 q 非常數（S10）。它是有限的端點商，沒有取極限。",
        "en": "The other misconception is to call the average rate a derivative. Starting from x₀=1, x₁=4 gives 2 and x₁=2 gives 6 (S11), so it is not a quantity of one point; the symmetric endpoints 1, 5 give 0 although q is not constant (S10). It is a finite endpoint quotient; no limit is taken."
      },
      {
        "title": "端點要成對一起對調 / Swap the endpoints as pairs",
        "zh": "(q(4)−q(1))/(1−4)=6/(−3)=−2 是錯的：只把分母的次序反過來，分子沒有一起反轉。兩者同時反轉才保持商不變（S04）。",
        "en": "(q(4)−q(1))/(1−4)=6/(−3)=−2 is wrong: only the order in the denominator was reversed, not in the numerator. Reversing both together keeps the quotient unchanged (S04)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不定義也不計算導數，只證明平均變化率依賴兩端點；與導數的關係需另行處理。",
        "en": "This guide neither defines nor computes derivatives; it only proves that the average rate depends on both endpoints. Its relation to the derivative is a separate matter."
      },
      {
        "zh-Hant": "G2 的 f 只透過兩個輸出值給出，本稿沒有使用 f 的其他任何資訊。",
        "en": "The f of G2 is given only through two outputs, and no other information about f is used."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "9fc30d6fb2c4750764240925dfd0912b58ac51b42cb81e261c6e4f836b55fb55",
    "original_source_sha256": "c2b90bff75e0c55cc21b498fdb786490c14ae05bee992190158da4a7dd811bad",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M016"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}