Can a point of inflection exist where the first derivative is undefined?

Question

In Thomas's Calculus book, an inflection point is defined as: "A point where the graph of a function has a tangent line and where the concavity changes". Then the definition is followed by this example: $f(x)= x^{1/3}$ where $x=0$ is a point of inflection. But the 1st derivative is undefined at $x=0$ (i.e. there is no tangent line), which contradicts the definition. How can this be possible?

Answer 1

There is no contradiction. A graph of a function can have a tangent without being differentiable. Note that $x = 0$ (the $y$-axis) is a tangent to $f$ at $(0,0)$.

Answer 2

The notion of inflection point is a geometric idea that should not be restricted to graphs $y=f(x)$. We are talking about regular curves $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .\tag{1}$$ Regular means that $\gamma$ is continuously differentiable, and that $\bigl(x'(t),y'(t)\bigr)\ne(0,0)$ on $[a,b]$. The tangent vector $\bigl(x'(t),y'(t)\bigr)$ has for all $t$ a well defined argument (polar angle) $$\theta(t):={\rm arg}\bigl(x'(t),y'(t)\bigr)\ .$$ To be precise, this $\theta(t)$ is only defined up to multiples of $2\pi$, but this does not need to bother us here. If $x'(t)>0$ then we may put $$\theta(t)=\arctan{y'(t)\over x'(t)}\ .$$

Now let $t_0$ be an interior point of the interval $[a,b]$. Then $\gamma$ has an inflection point at $P_0:=\gamma(t_0)$ if the function $t\mapsto\theta(t)$ is strictly increasing for $t\leq t_0$ and is strictly decreasing for $t\geq t_0$, or conversely.

If $\gamma$ is presented as graph $y=f(x)$ of a $C^2$-function then the above definition translates into the well known condition that $f''$ should change sign at $x_0$. In order to do Thomas' example correctly you should present the cubic parabola in the form $x=f(y):=y^3$ or by means of the parametrization $y\mapsto(y^3,y)$.

Answer 3

There is a problem in what you are claiming to be a definition,tangent may exist but derivative not at a point on a curve.