Goal and scope
For f=[3,−4,6] (f(x)=3−4x+6x²): prove from the definition the derivative formula for polynomials of degree at most four; build the representative antiderivative F0=[0,3,−2,2] and verify F0'=f; prove that F0+C is an antiderivative for every real C and that, among polynomials with at most five coefficients, every antiderivative is F0+C; check the candidates G=[7,3,−2,2] (accept), H=[0,3,−4,2] (reject) and K=[0,3,−3,8/3] (reject although it matches at two test points); evaluate F0(2)+C with C=−5; and explain why an antiderivative is not an inverse function.
Works only inside the MKO's polynomial model on all of ℝ: the concrete input f has at most four integer coefficients and exercise candidates such as G,H,K have at most five rational coefficients in ascending lists; the general family F0+C and its coefficient proof allow real C and real polynomial coefficients. 'Differ only by a constant' is proved by coefficients within the polynomial model only; the version for general differentiable functions on an interval needs the mean value theorem and is not proved here. It does not change the canonical MKO or its English edition.
Main derivation
S01 · definition
Before
F\ \text{antiderivative of}\ f\ \text{on}\ \mathbb{R}After
F\ \text{differentiable on}\ \mathbb{R}\ \land\ \forall x\in\mathbb{R}:\ F'(x)=f(x)Conditions: none
Rule: mko_antideriv · Uses: G1
Leaf basis: mko_antideriv + def_unfold
By the MKO: an antiderivative F must be differentiable on the interval (here all of ℝ), with derivative equal to f everywhere. This is a condition about derivatives and has nothing to do with inverse functions or reciprocals of derivatives.
S02 · equivalent
Before
f=[3,-4,6]
After
f(x)=3-4x+6x^2
Conditions: none
Rule: def_unfold · Uses: G2, G1
Leaf basis: def_unfold
Ascending reading: constant 3, coefficient −4 of x, coefficient 6 of x².
S03 · implies
Before
p=[c_0,c_1,c_2,c_3,c_4]
After
p'=[c_1,2c_2,3c_3,4c_4]\ \text{on}\ \mathbb{R}Conditions: none
Rule: L_poly_deriv4 · Uses: mko_derivative_def
Leaf basis: mko_derivative_def + distributive + abs_rules + order_basics
By Lemma L_poly_deriv4, a polynomial of degree at most four is differentiable everywhere with derivative list [c1,2c2,3c3,4c4]; the constant c0 does not appear in it. The lemma is proved from the definition in S03.1 to S03.3. Every polynomial differentiated here has degree at most four.
S04 · definition
Before
a_k\ \mapsto\ \tfrac{a_k}{k+1}\ \text{as the coefficient of}\ x^{k+1}After
F_0=[0,\ \tfrac31,\ \tfrac{-4}{2},\ \tfrac63]=[0,3,-2,2]Conditions: none
Rule: mko_antideriv · Uses: S02, mko_antideriv
Leaf basis: mko_antideriv + frac_rules + numeric_evaluation
By the MKO's representative formula, the coefficient a_k of x^k becomes the coefficient a_k/(k+1) of x^(k+1), with constant term 0. 3/1=3, −4/2=−2, 6/3=2 (N4, N5), so F0=3x−2x²+2x³. Coefficients are written as reduced fractions with positive denominators.
S05 · implies
Before
F_0=[0,3,-2,2]
After
F_0'=[3,\ 2\cdot(-2),\ 3\cdot 2]=[3,-4,6]=f;\ F_0\ \text{is an antiderivative}Conditions: none
Rule: L_poly_deriv4 · Uses: S03, S04, S02, S01
Leaf basis: L_poly_deriv4 + mko_antideriv + numeric_evaluation
By the lemma, F0 is differentiable everywhere with derivative list [3,−4,6], that is 3−4x+6x²=f(x) at every x (N6 is a polynomial identity). By S01, F0 is an antiderivative of f on ℝ.
S06 · implies
Before
F_0+C=[C,3,-2,2],\ C\in\mathbb{R}After
(F_0+C)'=[3,-4,6]=f\ \text{for every}\ CConditions: none
Rule: L_poly_deriv4 · Uses: S03, S05
Leaf basis: L_poly_deriv4 + forall_intro
The list of F0+C differs only in the constant term; the lemma's derivative list does not contain c0, so the derivative is still [3,−4,6]=f. As C is arbitrary, generalization gives that the whole family F0+C consists of antiderivatives. Giving one antiderivative F0 differs from giving all of them.
S07 · implies
Before
G=[7,3,-2,2]
After
G=F_0+7,\ G'=f:\ \text{accept}Conditions: none
Rule: L_poly_deriv4 · Uses: G3, S06
Leaf basis: L_poly_deriv4 + forall_elim
G differs from F0 only in the constant term (7 versus 0), so G=F0+7; by S06 with C=7, G'=f, and G is accepted. A candidate's constant term does not affect acceptance.
S08 · implies
Before
H=[0,3,-4,2]
After
H'=[3,-8,6]\neq[3,-4,6]=f:\ \text{reject}Conditions: none
Rule: L_poly_deriv4 · Uses: G3, S03, S02
Leaf basis: L_poly_deriv4 + quantifier_negation + numeric_evaluation
By the lemma, H'=[3,2·(−4),3·2]=[3,−8,6]. The coefficient at index 1 (counting from 0) is −8≠−4, so H'−f=−4x, not the zero polynomial; for example at x=1, H'(1)=1 and f(1)=5 (N7, N8). H is rejected.
S09 · implies
Before
P=[p_0,p_1,p_2,p_3,p_4],\ P'=f\ \text{on}\ \mathbb{R}After
P=F_0+p_0
Conditions: none
Rule: L_same_deriv · Uses: S03, S04
Leaf basis: poly_coeff_identity + frac_rules + mul_inverse_nonzero
By Lemma L_same_deriv, among polynomials with at most five coefficients, every antiderivative differs from F0 only in the constant term. This is the MKO's 'identical derivative coefficients force identical positive-degree coefficients'. The lemma is proved in S09.1 and S09.2.
S10 · equivalent
Before
F_0(2)+C,\ C=-5
After
(3\cdot 2-2\cdot 2^2+2\cdot 2^3)+(-5)=14-5=9
Conditions: none
Rule: numeric_evaluation · Uses: S04, G3
Leaf basis: numeric_evaluation
F0(2)=6−8+16=14 (N9), and adding C=−5 gives 9 (N10). Differentiation removes C, but evaluation does not: both C and x are genuine inputs and C may not be dropped just because the derivative does not contain it.
S11 · implies
Before
K=[0,3,-3,\tfrac83]
After
K'=[3,-6,8];\ K'(0)=f(0),\ K'(1)=f(1),\ K'(2)=23\neq 19=f(2):\ \text{reject}Conditions: none
Rule: L_poly_deriv4 · Uses: G3, S03, S02
Leaf basis: L_poly_deriv4 + quantifier_negation + numeric_evaluation
By the lemma, K'=[3,2·(−3),3·(8/3)]=[3,−6,8]. K'−f=2x²−2x=2x(x−1) (N14), which vanishes at x=0 and x=1, so sampling only those points would mislead (N11, N12); but it is not the zero polynomial, and at x=2, K'(2)=23 while f(2)=19 (N13). Comparing coefficients (−6≠−4) rejects K directly.
S12 · implies
Before
5x^0,\ 0
After
5x^0=5\ (\text{also at}\ x=0)\ \to\ [0,5];\ 0\ \to\ [0]Conditions: none
Rule: mko_antideriv · Uses: S03, mko_antideriv
Leaf basis: mko_antideriv + L_poly_deriv4 + power_laws
Monomial boundary: in polynomials x^0 is always 1, including at x=0, so 5x^0 is the constant 5 and its representative antiderivative is 5x, that is [0,5] (derivative [5] by the lemma). With coefficient c=0 the function is zero, and its representative antiderivative is normalized to [0], without meaningless higher zero terms; its family of antiderivatives is all constants.
S13 · implies
Before
p(x)=2x
After
\text{antiderivative}\ x^2;\ \text{inverse}\ y\mapsto\tfrac y2;\ 1/p'=\tfrac12:\ \text{three different objects}Conditions: none
Rule: mko_antideriv · Uses: S03, S01
Leaf basis: mko_antideriv + L_poly_deriv4 + quantifier_negation
For p(x)=2x: an antiderivative is x² ([0,0,1], whose derivative is [0,2]=p by the lemma); the inverse function is y↦y/2; the reciprocal of the derivative is 1/2. At input 4, x² gives 16 while the inverse gives 2 (N15). The three are entirely different: an antiderivative is neither an inverse function nor a reciprocal of the derivative.
S14 · implies
Before
\text{S05, S06, S09}After
\{P:\ P'=f,\ P\ \text{at most five coefficients}\}=\{F_0+C:\ C\in\mathbb{R}\}Conditions: none
Rule: set_extensionality · Uses: S05, S06, S09
Leaf basis: set_extensionality + subset_def
S06 gives ⊇ (every F0+C is an antiderivative) and S09 gives ⊆ (every such P is F0+p0). By set extensionality the sets are equal: in the polynomial model, the antiderivatives of f are exactly the constant family F0+C.
Substeps and lemmas
S03.1 · equivalent
Before
\frac{p(x+h)-p(x)}{h}\ (h\neq 0)After
D+hR,\ D=c_1+2c_2x+3c_3x^2+4c_4x^3,\ R=c_2+c_3(3x+h)+c_4(6x^2+4xh+h^2)
Conditions: h\neq 0
Rule: distributive · Uses: distributive, power_laws, frac_rules
Leaf basis: distributive + power_laws + frac_rules + mul_inverse_nonzero
Expand power by power: (x+h)²−x²=h(2x+h), (x+h)³−x³=h(3x²+h(3x+h)), (x+h)⁴−x⁴=h(4x³+h(6x²+4xh+h²)) (N1, N2, N3). Multiplying by the coefficients and adding, with c0 cancelling, gives p(x+h)−p(x)=h(D+hR); for h≠0 divide by h.
S03.2 · implies
Before
0<|h|\le 1
After
|R|\le M=|c_2|+|c_3|(3|x|+1)+|c_4|(6x^2+4|x|+1)
Conditions: |h|\le 1
Rule: abs_rules · Uses: S03.1, abs_rules
Leaf basis: abs_rules + order_basics + square_nonneg
Triangle inequality and |uv|=|u||v|, then |h|≤1 and h²≤1: |3x+h|≤3|x|+1 and |6x²+4xh+h²|≤6x²+4|x|+1. M depends only on x and the coefficients.
S03.3 · implies
Before
\varepsilon>0,\ \delta=\min\left(1,\tfrac{\varepsilon}{M+1}\right)After
0<|h|<\delta\Rightarrow|(D+hR)-D|\le|h|M<\varepsilon;\ p'(x)=D
Conditions: M+1>0
Rule: mko_derivative_def · Uses: S03.1, S03.2, mko_derivative_def
Leaf basis: mko_derivative_def + forall_intro + order_basics + abs_rules
If 0<|h|<δ then |h|<1 and S03.2 applies: |hR|≤|h|M≤|h|(M+1)<δ(M+1)≤ε. As ε is arbitrary, the definition gives p'(x)=D. As x is arbitrary, p is differentiable everywhere. S03.1 to S03.3 prove Lemma L_poly_deriv4.
S09.1 · implies
Before
P'(x)=f(x)\ \forall x
After
[p_1,2p_2,3p_3,4p_4]=[3,-4,6,0]
Conditions: none
Rule: poly_coeff_identity · Uses: S03, S02, poly_coeff_identity
Leaf basis: poly_coeff_identity + def_unfold
By S03, P' has list [p1,2p2,3p3,4p4]; f padded with a trailing zero is [3,−4,6,0] (the same polynomial). They agree at every x, so by poly_coeff_identity the coefficients agree term by term.
S09.2 · implies
Before
kp_k=a_{k-1}\ (k=1,2,3,4)After
p_1=3,\ p_2=-2,\ p_3=2,\ p_4=0;\ P=[p_0,3,-2,2]=F_0+p_0
Conditions: k\neq 0
Rule: frac_rules · Uses: S09.1, frac_rules
Leaf basis: frac_rules + mul_inverse_nonzero + numeric_evaluation
k=1, 2, 3, 4 are all nonzero, so we may divide: p1=3/1=3, p2=−4/2=−2, p3=6/3=2, p4=0/4=0. So the nonconstant coefficients of P are exactly those of F0, P=F0+p0, and the constant p0 is free. S09.1 and S09.2 prove Lemma L_same_deriv.