Inflection with vertical tangent

The third case: the one-sided derivatives are both infinite, but with the same sign. The tangent exists and is unique, but vertical — so it has no finite gradient.

Definition — Inflection with vertical tangent

$x_0$ is an inflection with vertical tangent of $f$ if both one-sided derivatives tend to $\pm\infty$ with the same sign: $f'_\pm(x_0) = +\infty$ or $f'_\pm(x_0) = -\infty$. The tangent exists and is unique, but it is vertical (infinite gradient).

Example — $f(x) = \sqrt[3]{x}$ at $x_0 = 0$

$f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$. Both as $x\to 0^-$ and as $x\to 0^+$, $\sqrt[3]{x^2}\to 0^+$, so $f'_\pm(0) = +\infty$. Both positive: inflection with vertical tangent (the curve crosses the $y$-axis with the $y$-axis itself as tangent).

At $x_0=0$ the tangent of $\sqrt[3]{x}$ is unique and vertical (it coincides with the $y$-axis).

Warning — Continuity is required

The classification presupposes that $f$ is continuous at $x_0$. If the function has a jump (first-kind discontinuity) at $x_0$, one does not speak of differentiability at $x_0$: non-differentiability is already a consequence of the discontinuity.

Links

Topics: Derivatives
Concepts: One-sided derivative · Non-differentiable point · Vertical tangent
Methods: Non-differentiable point vertical tangent
Skills: Calculating limits