What are non-differentiable points
In the entry on derivatives, we saw that if a function $f(x)$ is differentiable at a point $c$, then the function is continuous at that point. However, there are cases where a function is continuous at $c$ but not differentiable. For a function continuous at $c$, non-differentiability occurs when:
- The right-hand and left-hand limits of the difference quotient exist and are finite but are not equal, that is, $f_{-}'(c) \neq f_{+}'(c).$
- At least one of the two one-sided limits of the difference quotient is infinite.
- At least one of the two one-sided limits of the difference quotient does not exist, neither as a finite value nor as an infinite one.
The first two situations give rise to the three classical types of non-differentiable points, which we discuss below. The third situation, in which the difference quotient oscillates without approaching any value, is examined in a separate section.
Inflection points with vertical tangent
An inflection point is a point where the concavity of a function changes. When at such a point the one-sided limits of the difference quotient are both infinite with the same sign, the tangent line to the graph is parallel to the $y$-axis, and we obtain an inflection point with vertical tangent. If the function is increasing through the point, both one-sided derivatives equal $+\infty$:
$$f_{-}'(c) = f_{+}'(c) = +\infty$$
If the function is decreasing through the point instead, the common value is $-\infty$:
$$f_{-}'(c) = f_{+}'(c) = -\infty$$
In both cases the slopes of the secant lines grow without bound in absolute value as the second point approaches $c$, while the change of concavity forces the graph to cross its vertical tangent at the point.
As an example, consider the irrational function $f(x) = \sqrt[3]{x-1}$ at the point $c = 1$. The function is continuous on all of $\mathbb{R}$, and for $x \neq 1$ its derivative is:
$$f'(x) = \frac{1}{3\sqrt[3]{(x-1)^2}}$$
Since the quantity $\sqrt[3]{(x-1)^2}$ is positive and tends to zero as $x \to 1$ from either side, both one-sided limits of the derivative are $+\infty$:
$$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) = +\infty$$
For $x<1$, the function is concave upward, while for $x>1$ it is concave downward, so the point $(1, 0)$ is an inflection point with vertical tangent.
The function $g(x) = \sqrt[3]{1-x}$ exhibits the opposite behavior at the same point, since it is decreasing on all of $\mathbb{R}$ and both one-sided derivatives at $c = 1$ equal $-\infty$.
Cusps
In the case of cusps, the right-hand and left-hand limits of the difference quotient are infinite and have opposite signs. For a downward-pointing cusp we have:
$$f_{-}'(c) = -\infty \quad \text{and} \quad f_{+}'(c) = +\infty$$
If the cusp faces upward instead, the signs are reversed:
$$f_{-}'(c) = +\infty \quad \text{and} \quad f_{+}'(c) = -\infty$$
A typical example is the power function $f(x) = x^{2/3} = \sqrt[3]{x^2}$ at the point $c = 0$. The function is continuous on all of $\mathbb{R}$, and for $x \neq 0$ its derivative is:
$$f'(x) = \frac{2}{3\sqrt[3]{x}}$$
The cube root $\sqrt[3]{x}$ is negative for $x < 0$ and positive for $x > 0$, so the one-sided limits of the derivative are:
$$\lim_{x \to 0^-} f'(x) = -\infty \quad \text{and} \quad \lim_{x \to 0^+} f'(x) = +\infty$$
The graph descends with increasingly steep slope to the left of the origin and rises with increasingly steep slope to the right, forming a downward-pointing cusp at $(0, 0)$, which is also a global minimum point of the function.
Corners
A corner occurs when the left-hand derivative and the right-hand derivative are different and at least one of the two is finite. At a corner point, the graph admits two distinct one-sided tangent lines at the same point:
$$f_{-}'(c) \neq f_{+}'(c)$$
The simplest example is provided by the absolute value. Consider the function $f(x) = |x - 2|$ at the point $c = 2$. For $x > 2$ the function coincides with $x - 2$, whose derivative is $1$, while for $x < 2$ it coincides with $2 - x$, whose derivative is $-1$. The one-sided derivatives at $c = 2$ are therefore:
$$f_{-}'(2) = -1 \quad \text{and} \quad f_{+}'(2) = 1$$
Both one-sided derivatives exist and are finite, but they are different, so the function is continuous at $x = 2$ without being differentiable there. The point $(2, 0)$ is a corner, and it is also a global minimum point of the function.
A corner can also combine a finite one-sided derivative with an infinite one. Consider the following piecewise function:
$$ f(x) = \begin{cases} x & x \leq 0 \\[6pt] \sqrt{x} & x > 0 \end{cases} $$
The function is continuous at $c = 0$, since both pieces vanish at the origin. For $x < 0$ the function coincides with $x$, so $f_{-}'(0) = 1$. For $x > 0$ the difference quotient at the origin is:
$$\frac{\sqrt{x} - 0}{x - 0} = \frac{1}{\sqrt{x}}$$
This quantity tends to $+\infty$ as $x \to 0^+$, so $f_{+}'(0) = +\infty$. The graph approaches the origin with slope $1$ from the left and with a vertical one-sided tangent from the right, and the point $(0, 0)$ is still classified as a corner because the two one-sided tangent lines are distinct and one of the one-sided derivatives is finite.
Oscillating difference quotients
The limit of the difference quotient may also fail to exist altogether, without being infinite. In this case the point of non-differentiability belongs to none of the three types described above. Consider the function defined by:
$$ f(x) = \begin{cases} x \sin\left(\dfrac{1}{x}\right) & x \neq 0 \\[6pt] 0 & x = 0 \end{cases} $$
The bound $|f(x)| \leq |x|$ holds for every $x$, so by the squeeze theorem the function is continuous at $c = 0$. The difference quotient at the origin is:
$$\frac{f(x) - f(0)}{x - 0} = \sin\left(\frac{1}{x}\right)$$
As $x \to 0$, this expression oscillates between $-1$ and $1$ infinitely many times, so neither one-sided limit exists, not even as an infinite value. The function is therefore continuous but not differentiable at the origin, and the point is neither a corner, nor a cusp, nor an inflection point with vertical tangent.
A criterion based on the limit of the derivative
How can we verify the differentiability of a function without relying on the limit of its difference quotient?
In general, let $f(x)$ be a function continuous on an interval $[a,b]$ and differentiable on that interval, except possibly at the point $x_0 \in [a,b]$. If the limits $\lim_{x \to x_0^-} f'(x)$ and $\lim_{x \to x_0^+} f'(x)$ exist, finite or infinite, then:
$$f_{-}'(x_0) = \lim_{x \to x_0^-} f'(x) \quad \text{and} \quad f_{+}'(x_0) = \lim_{x \to x_0^+} f'(x)$$
If the two limits are equal to the same finite value $\ell \in \mathbb{R}$, that is:
$$\lim_{x \to x_0^-} f'(x) = \lim_{x \to x_0^+} f'(x) = \ell$$
then the function is differentiable at $x_0$, and it follows that $f'(x_0) = \ell$. This criterion is a consequence of Lagrange's Theorem, which allows each one-sided limit of the derivative to be transferred to the corresponding one-sided limit of the difference quotient.
When $f$ is differentiable on the whole interval, including $x_0$, the two one-sided limits of $f'$ cannot exist and differ. A derivative has no jump discontinuity, by Darboux's Theorem, so whenever both one-sided limits are finite they are forced to agree, and $f'$ is continuous at $x_0$. A corner, where the one-sided derivatives are finite and distinct, can arise only at a point where the function is not differentiable to begin with.
The hypothesis of continuity at $x_0$ is essential. If the function has a jump at $x_0$, the one-sided limits of $f'(x)$ may exist and coincide even though the function is not differentiable, and not even continuous, at the point.
The criterion provides a sufficient condition only. A function can be differentiable at $x_0$ even when the limit of $f'(x)$ as $x \to x_0$ does not exist. The function defined by $f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ is differentiable at the origin, since its difference quotient $x\sin(1/x)$ tends to $0$, so that $f'(0) = 0$. For $x \neq 0$, however, the derivative is:
$$f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$$
The first term tends to $0$, while the second oscillates between $-1$ and $1$, so $f'(x)$ has no limit as $x \to 0$. The failure of the criterion in this direction does not allow any conclusion about non-differentiability.