The Derivative and the Mean Value Theorems: Examples and Counterexamples

Numerals below denote canonical naturals of $\mathbb{R}$: $3$ is $\iota (3)$, $12$ is $\iota (12)$, and so on (The canonical natural $\iota (n)=n\cdot 1_F$ of a field). Powers are those of Integer powers $a^m$.

Claim 1. Let $p:\mathbb{R}\to \mathbb{R}$ be given by

$p(x)=\iota (3)x^4-\iota (5)x+\iota (2)\mathrm{.}$

Then $p$ is differentiable at every $c\in \mathbb{R}$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set) and

$p^{\mathrm{\prime }}(c)=\iota (12)c^3-\iota (5)\mathrm{.}$

Claim 2. Put $A_0:=\{x\in \mathbb{R}:x-1\mathrm{\ne }0\}=\mathbb{R}\setminus \{1\}$ and let $r:A_0\to \mathbb{R}$ be given by $r(x)=(x^2+1)\mathrm{/}(x-1)$. Then every $c\in A_0$ is a limit point of $A_0$, $r$ is differentiable at $c$ as a function on $A_0$, and

$r^{\mathrm{\prime }}(c)=\frac{c^2-\iota (2)c-1}{(c-1{)}^2}\mathrm{.}$

Nothing here is new: both computations are readings of Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$ on top of For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term. They are written out because the two places a computation of this kind goes wrong are the constant term, whose derivative is $0$ and not $\iota (0)c^{-1}$, and the domain of the quotient, which is not $\mathbb{R}$.

• The constant term is where the index trap sits. Written informally, the derivative of $\iota (2)x^0$ "is" $\iota (2)\iota (0)x^{-1}$, and $x^{-1}$ is undefined at $0$. Claim 1 of For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term exists precisely so that the constant case is handled by its own statement, and the answer there is $0$ on the whole line.

• The quotient lives on $\mathbb{R}\setminus \{1\}$, not on $\mathbb{R}$. The function $r$ is not defined at $1$, and no derivative of it there is asserted or could be. Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$ states its quotient case on the set where the denominator does not vanish for exactly this reason, and it also supplies the fact that the smaller set still has $c$ as a limit point, without which the derivative there would not be a defined symbol.

• Reading the numerals. $\iota (12)$ is what "$12$" means as an element of $\mathbb{R}$; the equality $\iota (3)\iota (4)=\iota (12)$ is a lemma (Canonical naturals are positive and strictly increasing) and not an act of arithmetic on the page. Every numeral in this library is such an image, and where a computation multiplies two of them the lemma is what licenses collapsing the product.

Refuted claim: if $A\subseteq \mathbb{R}$, if $f:A\to \mathbb{R}$ is continuous at a point $c\in A$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) and if $c$ is a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), then $f$ is differentiable at $c$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set).

This is the converse of A function differentiable at $c$ is continuous at $c$, and it is false. The witness is $f(x)=\mathrm{\mid }x\mathrm{\mid }$ on $A=\mathbb{R}$ at $c=0$: a single corner is enough, and the failure is visible in one line, the difference quotient taking the value $1$ to the right of $0$ and $-1$ to the left.

• The failure is one-sided in a precise sense. Both one-sided limits of the difference quotient exist; they simply disagree. So this is not a function whose difference quotients oscillate or blow up, and the restriction of $f$ to $[0,\mathrm{\infty })$ is differentiable at $0$ with derivative $1$, as The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set records when it observes that differentiability is a property of the pair (function, domain).

• What this says about A function differentiable at $c$ is continuous at $c$. That implication is strict: continuity is genuinely weaker than differentiability, and this witness shows the gap opens at a single point of an otherwise unremarkable function. Nothing here suggests the gap is small in any other sense; how large the set of non-differentiability of a continuous function can be is not a question this page can pose.

• Why the argument needs If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree and not merely two computations. Two different one-sided values do not by themselves contradict anything until one knows that a two-sided limit would have to agree with both, and that is exactly what the cited theorem supplies, under the hypothesis that $0$ is approached from both sides inside the domain.

Let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=x^3$ (Integer powers $a^m$), with $\iota$ the canonical natural of The canonical natural $\iota (n)=n\cdot 1_F$ of a field.

Claim 1. $f$ is differentiable at every $c\in \mathbb{R}$ with $f^{\mathrm{\prime }}(c)=\iota (3)c^2$, and $f^{\mathrm{\prime }}(0)=0$.

Claim 2. $f$ is increasing on $\mathbb{R}$, in the strict sense of Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences.

Claim 3. So the hypothesis "$f^{\mathrm{\prime }}>0$ at every interior point" of claim 2 of On an interval $I$, for $f$ continuous on $I$ and differentiable at every interior point: $f^{\mathrm{\prime }}\ge 0$ throughout gives $f$ nondecreasing, $f^{\mathrm{\prime }}>0$ gives $f$ increasing, $f^{\mathrm{\prime }}\le 0$ and $f^{\mathrm{\prime }}<0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{\mathrm{\prime }}\ge 0$ and a nonincreasing $f$ has $f^{\mathrm{\prime }}\le 0$ wherever it is differentiable, and no strict converse is claimed is sufficient but not necessary for a function on an interval to be increasing; and the converse recorded there, claim 5, which gives only $f^{\mathrm{\prime }}\ge 0$, cannot be strengthened to $f^{\mathrm{\prime }}>0$.

Claim 4. $f$ is continuous and injective on $\mathbb{R}$, so it has a continuous inverse $g$ on $f[\mathbb{R}]$ (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c\in I$ with $f^{\mathrm{\prime }}(c)\mathrm{\ne }0$, then the inverse $g$ is differentiable at $f(c)$ with $g^{\mathrm{\prime }}(f(c))=1\mathrm{/}{f}^{\mathrm{\prime }}(c)$; and if $f^{\mathrm{\prime }}(c)=0$ then $g$ is not differentiable at $f(c)$); and since $f^{\mathrm{\prime }}(0)=0$, that inverse is not differentiable at $f(0)=0$.

Claims 1 and 2 are established in the refutation of FALSE: if $f^{\mathrm{\prime }}(c)=0$ then $f$ is not increasing on any interval containing $c$ and are quoted here; claims 3 and 4 are the two consequences worth drawing from them.

• The set $f[\mathbb{R}]$ is not identified here, and nothing needs it to be. Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c\in I$ with $f^{\mathrm{\prime }}(c)\mathrm{\ne }0$, then the inverse $g$ is differentiable at $f(c)$ with $g^{\mathrm{\prime }}(f(c))=1\mathrm{/}{f}^{\mathrm{\prime }}(c)$; and if $f^{\mathrm{\prime }}(c)=0$ then $g$ is not differentiable at $f(c)$ states its conclusion about the inverse on $f[I]$, whatever that set is; that $f[\mathbb{R}]=\mathbb{R}$, so that $g$ is the cube root on the whole line, would need a surjectivity argument this item does not make and does not use.

• Two different false readings, one witness. That $f^{\mathrm{\prime }}(c)=0$ forbids strict increase is refuted by claim 2; that $f^{\mathrm{\prime }}(c)=0$ marks a local extremum is refuted by the same fact, since an increasing function has no local extremum at an interior point of its interval. The second reading is the converse of Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$, then $f^{\mathrm{\prime }}(c)=0$, and this page records it here rather than as a separate false statement.

• Where the inverse fails, and why it is not surprising. Claim 4 is not a defect of the inverse rule but a theorem: wherever the derivative of an injective continuous function vanishes, the inverse cannot be differentiable, because the chain rule would then give the identity a derivative of $0$. The cube root at $0$ is the standard picture of that, a vertical tangent, and it is proved here without any picture.

Numerals denote canonical naturals of $\mathbb{R}$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field) and powers are those of Integer powers $a^m$.

Claim 1. Let $F:\mathbb{R}\to \mathbb{R}$ be $F(x)=(x^2+1{)}^5$. Then $F$ is differentiable at every $c\in \mathbb{R}$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set) and

$F^{\mathrm{\prime }}(c)=\iota (10)c(c^2+1{)}^4\mathrm{.}$

Claim 2. For $b\in \mathbb{R}$ the Carathéodory factor (Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $\phi :A\to \mathbb{R}$, continuous at $c$, with $f(x)-f(c)=\phi (x)(x-c)$ for every $x\in A$, and then $\phi$ is unique and $\phi (c)=f^{\mathrm{\prime }}(c)$) of $y\mapsto y^5$ at $b$ is the polynomial function

$\phi (y)={\sum }_{k<5}{b}^k{y}^{4-k},$

which satisfies $y^5-b^5=\phi (y)(y-b)$ for every $y\in \mathbb{R}$, is continuous at $b$, and has $\phi (b)=\iota (5)b^4$.

Claim 3. Let $G:\mathbb{R}\to \mathbb{R}$ be $G(x)=((\iota (3)x-1{)}^2+2{)}^3$. Then $G$ is differentiable at every $c\in \mathbb{R}$ and

$G^{\mathrm{\prime }}(c)=\iota (18)(\iota (3)c-1)((\iota (3)c-1{)}^2+2{)}^2\mathrm{.}$

Claim 2 is included because it makes the mechanism of the chain rule visible: the factor that the proof of The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$ takes from Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $\phi :A\to \mathbb{R}$, continuous at $c$, with $f(x)-f(c)=\phi (x)(x-c)$ for every $x\in A$, and then $\phi$ is unique and $\phi (c)=f^{\mathrm{\prime }}(c)$ is, for a power, an explicit polynomial, and no auxiliary case distinction is hidden inside it.