Average rate of change between distinct inputs: eligibility, swapping endpoints, the affine and quadratic formulas

Goal and scope

(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.

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.

Givens and symbols

• frac_rules
• distributive
• mul_inverse_nonzero
• neg_mul
• numeric_evaluation

• G1 — 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.
• G2 — Direct two-point exercise: the specified function f is given only through f(2)=7 and f(−4)=3; take x₀=2, x₁=−4.
• G3 — ℓ:ℝ→ℝ, ℓ(x)=−3x+7 (m=−3, b=7).
• G4 — q:ℝ→ℝ, q(x)=−2(x−3)²+5 (a=−2≠0, h=3, k=5).

• x_0,x_1 — the two input endpoints, x_0≠x_1
• y_0=f(x_0),\ y_1=f(x_1) — the corresponding outputs
• R(f;x_0,x_1)=\frac{y_1-y_0}{x_1-x_0} — the average rate of change
• \ell(x)=-3x+7 — the degree-one affine example
• q(x)=-2(x-3)^2+5 — the quadratic example

Rules and sources

• frac_rules · prerequisite — 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.
• distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
• mul_inverse_nonzero · prerequisite — 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.
• neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv.
• numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
• def_intro · prerequisite — Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable.
• add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
• add_comm · prerequisite — Commutativity of addition: u+v=v+u.
• mul_assoc · prerequisite — Associativity of multiplication: (uv)w=u(vw).
• mul_one · prerequisite — Multiplicative identity: 1·u=u.
• mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
• quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x).
• and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
• mko_rate · cited — 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. · mko-rate-of-change
• diff_squares · prerequisite — Difference of squares: for all reals u, v, u²−v²=(u−v)(u+v) (expand the right side by distributivity).
• L_swap · lemma — For x₀≠x₁: (y₀−y₁)/(x₀−x₁)=(y₁−y₀)/(x₁−x₀).
• L_linear_rate · lemma — For f(x)=mx+b and x₀≠x₁: (f(x₁)−f(x₀))/(x₁−x₀)=m.
• L_quad_rate · lemma — For q(x)=a(x−h)²+k and x₀≠x₁: (q(x₁)−q(x₀))/(x₁−x₀)=a(x₀+x₁−2h).

Main derivation

f(x_0)=y_0,\ f(x_1)=y_1

R=\frac{y_1-y_0}{x_1-x_0}\ \ \text{only if}\ x_0\neq x_1

x_0=2,\ x_1=-4

x_1-x_0=-6\neq0

R=\frac{3-7}{-4-2}

\frac{2}{3}

\frac{y_0-y_1}{x_0-x_1}=\frac{7-3}{2-(-4)}

\frac{4}{6}=\frac{2}{3}

\frac{(mx_1+b)-(mx_0+b)}{x_1-x_0}

m

\ell(-2)=13,\ \ell(4)=-5

\frac{-5-13}{4-(-2)}=\frac{-18}{6}=-3=m

\frac{q(x_1)-q(x_0)}{x_1-x_0}

a(x_0+x_1-2h)

q(1)=-3,\ q(4)=3

\frac{3-(-3)}{4-1}=\frac{6}{3}=2

a(x_0+x_1-2h)\ \text{at}\ a=-2,\ h=3,\ x_0=1,\ x_1=4

-2(1+4-6)=2

x_0=1,\ x_1=5=2h-1

R=-2(1+5-6)=0,\ \text{yet}\ q(3)=5\neq-3=q(1)

x_0=1,\ x_1=2

R=-2(1+2-6)=6\neq2

x_0=x_1=2

x_1-x_0=0:\ \text{ineligible, no value assigned}

\text{S02 to S12}

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}

Substeps and lemmas

\frac{3-7}{-4-2}

\frac{-4}{-6}

\frac{-4}{-6}

\frac{4}{6}=\frac{2}{3}

\frac{y_0-y_1}{x_0-x_1}

\frac{-(y_1-y_0)}{-(x_1-x_0)}=\frac{y_1-y_0}{x_1-x_0}

(mx_1+b)-(mx_0+b)

m(x_1-x_0)

\frac{m(x_1-x_0)}{x_1-x_0}

m

q(x_1)-q(x_0)

a\bigl[(x_1-h)^2-(x_0-h)^2\bigr]

(x_1-h)^2-(x_0-h)^2

(x_1-x_0)(x_0+x_1-2h)

\frac{a(x_1-x_0)(x_0+x_1-2h)}{x_1-x_0}

a(x_0+x_1-2h)

Reverse audits

• S13 · sufficient_goal_reduction · S03, S04, S05, S06, S07, S09, S10, S11, S12, and_intro_elim — 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.
• S12 · premise_audit · mko_rate, G1 — The eligibility condition itself excludes this case; no limit or derivative is claimed.
• S11 · premise_audit · S07, S09, numeric_evaluation — Two different values suffice to show the dependence on the second endpoint; the guide does not compute the derivative itself.
• S10 · premise_audit · S07, numeric_evaluation — One counterexample (q(3)≠q(1)) suffices to refute 'zero average rate ⇒ constant'.
• S09 · premise_audit · S07, numeric_evaluation — The formula is proved in S07; agreement of the two computations is only a check.
• S08 · premise_audit · mko_rate, numeric_evaluation — Eligibility 4≠1 holds; exact arithmetic.
• S07.3 · inverse_equivalence · mul_inverse_nonzero — Cancellation is reversible, given x₁−x₀≠0.
• S07.2 · inverse_equivalence · diff_squares, add_inverse — The difference of squares is an identity and holds both ways.
• S07.1 · inverse_equivalence · distributive, add_inverse — An identity; it holds both ways.
• S07 · inverse_equivalence · L_quad_rate — The lemma is proved by three equational sub-steps and holds both ways.
• S06 · premise_audit · S05, numeric_evaluation — The general conclusion comes from S05; the numbers are only a check.
• S05.2 · inverse_equivalence · mul_inverse_nonzero — Multiplying by (x₁−x₀) and by its reciprocal undo each other, given x₁−x₀≠0.
• S05.1 · inverse_equivalence · distributive, add_inverse — An identity; it holds both ways.
• S05 · inverse_equivalence · L_linear_rate — The lemma is an equation and holds both ways.
• S04.1 · inverse_equivalence · frac_rules, neg_mul — Both rules are identities whose only condition is a nonzero denominator.
• S04 · inverse_equivalence · L_swap — The lemma is an equation and holds both ways.
• S03.2 · inverse_equivalence · frac_rules — a/b=(ak)/(bk) for k≠0 holds both ways.
• S03.1 · inverse_equivalence · numeric_evaluation — Integer subtraction, checkable backwards.
• S03 · inverse_equivalence · frac_rules, numeric_evaluation — Each step rewrites the same rational number, so it holds both ways.
• S02 · premise_audit · S01, numeric_evaluation — Exact computation of the input difference and comparison with 0.
• S01 · premise_audit · G1, mko_rate — This is only the definition and its eligibility condition; for x₀=x₁ no value is defined.

Error controls and negative controls

Do not assign zero when the inputs coincide

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).

The average rate is not a derivative

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.

Swap the endpoints as pairs

(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).

• For x₀=x₁=2 the average rate of q is 0 → reject — The denominator is 0, so the pair is ineligible; 0/0 is not a number and may not be replaced by 0.
• (q(4)−q(1))/(1−4)=−2 is the average rate between 1 and 4 → reject — Only the denominator was reversed; the correct value is (q(4)−q(1))/(4−1)=2.
• q has average rate 0 between 1 and 5, so q is constant on [1,5] → reject — q(3)=5≠−3=q(1).
• 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 → reject — 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.

Result and limitations

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.

• 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.
• The f of G2 is given only through two outputs, and no other information about f is used.

Verification and source binding

21 declared checks for this guide; only the listed expressions are covered.

• source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
• adopted source SHA-256: 9fc30d6fb2c4750764240925dfd0912b58ac51b42cb81e261c6e4f836b55fb55
• original source SHA-256: c2b90bff75e0c55cc21b498fdb786490c14ae05bee992190158da4a7dd811bad
• Lean: NOT_RUN · independent AI: NOT_MEASURED

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