$x\mapsto x^3$ is increasing on $\mathbb{R}$ although its derivative vanishes at $0$, which is the witness for the false statement that a vanishing derivative forbids strict increase, and which makes its inverse non-differentiable at $0$

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.