PROOF 138

Prove.

The derivative can be 0 without an extremum

On the real numbers ℝ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

What you may use

A10 < a, 0 < b ⟹ 0 < ab
Order and multiplication
A2(xⁿ)′ = nxⁿ⁻¹
Derivative of a power (Proof 135)
A3f′>0 ⟹ increasing, f′<0 ⟹ decreasing
Sign of the derivative (Proof 136)
A4a < b ⟺ 0 < b − a
Order and subtraction
A50 < a ⟹ 0 < a/2 < a
Half of a positive number
A6(−a)b = −(ab)
Signs and multiplication
A7−(−a) = a
Sign of a sign
A8if a < b then not b ≤ a
Trichotomy of order
A9a × 0 = 0
Multiplying by 0 gives 0
A10ab = ba
Commutativity of multiplication
A11(ab)c = a(bc)
Associativity of multiplication
A12a + b = b + a
Commutativity of addition
A13a + 0 = a
Additive identity

D1|h|<δ ⟹ f(c+h)≤f(c)
Definition of a local maximum
T1x³ : f′(0)=0, not an extremum
Theorem 1
Ca sign change gives an extremum
Corollary — when it is an extremum

Nothing outside this list may be cited.

Definition D1

On an open interval containing c, if there is some δ > 0 with f(c+h) ≤ f(c) for every h with |h| < δ, then c is called a local maximum point of f. Reversing the inequality gives a local minimum point.

Proof.

Being neither a local maximum nor a local minimum means that however small δ is taken, both inequalities break.convention

∎

Theorem T1

f(x) = x³ has f′(0) = 0, yet 0 is neither a local maximum nor a local minimum point of f.

Proof.

(1)f′(x) = 3x² and f′(0) = 3 · (0 · 0) = 0.A2, A9

(2)If 0 < x, then 0 < x · x, and again 0 < (x · x) · x = x³.A1, A11

(3)(−x)(−x) = −((−x)x) = −(−(x · x)) = x · x, and (x · x)(−x) = −(x · x · x).A6, A7, A10, A11

(4)Therefore (−x)³ = −(x³).(3), A11

(5)If x < 0, then 0 < −x, so by (2), 0 < (−x)³ = −(x³), and therefore x³ < 0.(2), (4), A4, A12, A13

(6)However small δ > 0 is taken, 0 < δ/2 < δ, so among h with |h| < δ there are δ/2 and −δ/2.A5, A4

(7)At h = δ/2, f(0+h) = h³ > 0 = f(0), so f(0+h) ≤ f(0) breaks. 0 is not a local maximum point.(2), (6), A8, A12, A13, A9, D1

(8)At h = −δ/2, f(0+h) = h³ < 0 = f(0), so f(0) ≤ f(0+h) breaks. 0 is not a local minimum point.(5), (6), A8, A12, A13, A9, D1

(9)Therefore f′(0) = 0 is not a sufficient condition for an extremum.(1), (7), (8)

∎

Corollary C

If f is continuous on an open interval containing c, differentiable except at c, and f′ > 0 to the left of c and f′ < 0 to the right, then c is a local maximum point of f.

Proof.

(1)Take δ > 0 so that the hypotheses hold on (c−δ, c+δ).assumption

(2)f′ > 0 on [c−δ, c], so f increases on that interval, and if c−δ ≤ x < c, then f(x) < f(c).(1), A3

(3)f′ < 0 on [c, c+δ], so f decreases on that interval, and if c < x ≤ c+δ, then f(x) < f(c).(1), A3

(4)Therefore f(c+h) ≤ f(c) for every h with |h| < δ, and c is a local maximum point.(2), (3), D1

(5)f(x) = x³ does not satisfy these hypotheses. By T1, f(x) < f(0) to the left of 0 and f(0) < f(x) to the right, so (2) and (3) cannot both hold.T1, (2), (3)

∎

Remark

Can A1 be removed from this proof?

T1 examines the sign of x³ on both sides of 0. Check what says that the product of three positives is positive.