When A Function Is Not Differentiable At A Point

In this section, we study how f'(x_0) might fail to exist. Here are a number of cases in which $f$ is not differentiable at a point.

Quick Reference

Cause of Non-Differentiability	Algebraic / Limit Reason	Geometric Appearance	Example
Discontinuity	$f (x)$ is not continuous at $x_{0}$	Jump, hole, or vertical asymptote	Heaviside step function $H (x)$ , $F (x) = 1 / x^{2}$
Corner / Cusp	f'_+(x_0) \neq f'_-(x_0)	Sharp corner or V-shape point	$y = \| x \|$ at $x = 0$
Vertical Tangent	$lim_{Δ x \to 0} \frac{Δ y}{Δ x} = \pm \infty$	Tangent line is vertical (slope is infinite)	$y = \sqrt[3]{x}$ at $x = 0$
Wild Oscillation	$lim_{Δ x \to 0} \frac{Δ y}{Δ x}$ does not exist	Rapid infinite oscillation near point	$f (x) = x \sin (1 / x)$ at $x = 0$

Discontinuity at $x = x_{0}$ :
When $f$ is discontinuous at $x = x_{0}$ . For example, if there is a jump in the graph of $f$ at $x = x_{0}$ , or we have $lim_{x \to x_{0}} f (x) = + \infty$ or $- \infty$ , the function is not differentiable at the point of discontinuity. For example, consider: H(x) = \begin{cases} 1 & \text{if } 0 \le x, \\ 0 & \text{if } x < 0. \end{cases} This function, which is called the Heaviside step function, is not continuous at $x = 0$ ; therefore H'(0) does not exist. Another example is $F (x) = 1 / x^{2}$ . Because $lim_{x \to 0} F (x) = + \infty$ , $F$ is not differentiable at $x = 0$ .

The graphs of $y = H (x)$ and $y = \frac{1}{x^{2}}$ show that they are discontinuous at $x = 0$ . Therefore, they are not differentiable there.
Corner or Cusp (f'_+(x_0) \neq f'_-(x_0)):
When f'_+(x_0) and f'_-(x_0) both exist but f'_+(x_0) \neq f'_-(x_0). In this case, there is a corner in the graph of $f$ . For instance, see the following figure.

The function is not differentiable wherever the graph has a corner or cusp.
Vertical Tangent Line:
When the tangent line is vertical. In this case, $lim_{Δ x \to 0} \frac{f (x_{0} + Δ x) - f (x_{0})}{Δ x} = + \infty or - \infty,$ or more generally \lim_{\Delta x\to 0} |f'(x)| = +\infty. For example, consider $f (x) = x^{1 / 3}$ . As you can see in the figure below, the tangent to its graph at $(0, 0)$ is vertical. This function is not differentiable at $x = 0$ , for: \begin{aligned} \left.\frac{d(\sqrt[3]{x})}{dx}\right|_{x=0} &= \lim_{\Delta x\to 0} \frac{\sqrt[3]{0 + \Delta x} - \sqrt[3]{0}}{\Delta x} \\ &= \lim_{\Delta x\to 0} \frac{\sqrt[3]{\Delta x}}{\Delta x} \\ &= \lim_{\Delta x\to 0} \frac{1}{\Delta x^{(1 - 1/3)}} \\ &= \lim_{\Delta x\to 0} \frac{1}{\sqrt[3]{(\Delta x)^2}} \\ &= +\infty. \end{aligned}

The tangent line to the graph of $y = \sqrt[3]{x}$ at $(0, 0)$ is vertical.
Oscillation (No Limit):
When the difference quotient $\frac{f (x + Δ x) - f (x)}{Δ x}$ approaches neither a number nor $\pm \infty$ as $Δ x \to 0$ . For example, consider: f(x) = \begin{cases} x\sin\frac{1}{x} & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases} This function is not differentiable (although it is continuous) at $x = 0$ , because: \begin{aligned} f'(0) &= \lim_{\Delta x\to 0} \frac{f(0 + \Delta x) - f(0)}{\Delta x} \\ &= \lim_{\Delta x\to 0} \frac{\Delta x\sin\frac{1}{\Delta x} - 0}{\Delta x} \\ &= \lim_{\Delta x\to 0} \sin\frac{1}{\Delta x} \end{aligned} does not exist. The graph of $f (x)$ is shown below.

Graph of y = \begin{cases} x\sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}.

Exercises

Show that the function $f (x) = | x - 3 |$ is continuous at $x_{0} = 3$ but is not differentiable at $x_{0} = 3$ .

Solution

Recall that the absolute value function $f (x) = | x - 3 |$ can be written as a piecewise function:

f(x) = \begin{cases} x - 3 & \text{if } x \ge 3, \\ -(x - 3) & \text{if } x < 3. \end{cases}

Step 1: Check Continuity at $x = 3$

Evaluating the limits from the left and right:

lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} - (x - 3) = 0,

lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} (x - 3) = 0,

and $f (3) = | 3 - 3 | = 0$ . Since $lim_{x \to 3} f (x) = f (3) = 0$ , $f$ is continuous at $x = 3$ .

Step 2: Calculate One-Sided Derivatives at $x = 3$

For the left-hand derivative f'_-(3):

\begin{aligned} f'_-(3) &= \lim_{\Delta x\to 0^-} \frac{f(3 + \Delta x) - f(3)}{\Delta x} \\ &= \lim_{\Delta x\to 0^-} \frac{-((3 + \Delta x) - 3) - 0}{\Delta x} \\ &= \lim_{\Delta x\to 0^-} \frac{-\Delta x}{\Delta x} \\ &= -1. \end{aligned}

For the right-hand derivative f'_+(3):

\begin{aligned} f'_+(3) &= \lim_{\Delta x\to 0^+} \frac{f(3 + \Delta x) - f(3)}{\Delta x} \\ &= \lim_{\Delta x\to 0^+} \frac{((3 + \Delta x) - 3) - 0}{\Delta x} \\ &= \lim_{\Delta x\to 0^+} \frac{\Delta x}{\Delta x} \\ &= 1. \end{aligned}

Step 3: Conclusion

Since f'_-(3) = -1 \neq 1 = f'_+(3), the left and right limits of the difference quotient are not equal. The graph of $f (x) = | x - 3 |$ forms a sharp V-shaped corner at $(3, 0)$ , so f'(3) does not exist.

Geometric Graph:

The figure below illustrates the V-shaped corner of $f (x) = | x - 3 |$ at $(3, 0)$ .

Determine whether the piecewise function:

f(x) = \begin{cases} x^2 + 1 & \text{if } x < 1, \\ 4 & \text{if } x = 1, \\ 2x & \text{if } x > 1 \end{cases}

is differentiable at $x = 1$ .

Solution

Before checking differentiability, we must check whether $f$ is continuous at $x = 1$ .

Step 1: Check Continuity at $x = 1$

Evaluating the one-sided limits as $x \to 1$ :

lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (x^{2} + 1) = 1^{2} + 1 = 2,

lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (2 x) = 2 (1) = 2.

Thus $lim_{x \to 1} f (x) = 2$ . However, the function value at $x = 1$ is defined as $f (1) = 4$ .

Since $lim_{x \to 1} f (x) = 2 \neq 4 = f (1)$ , the function has a removable discontinuity at $x = 1$ .

Step 2: Conclusion

Since differentiability requires continuity, any point of discontinuity automatically prevents the existence of a derivative. Therefore, f'(1) does not exist because $f$ is discontinuous at $x = 1$ .

Geometric Graph:

The figure below illustrates the removable discontinuity (hole at $(1, 2)$ and isolated point at $(1, 4)$ ).

Investigate the differentiability of $f (x) = x^{2 / 3}$ at $x = 0$ .

Solution

Notice that $f (0) = 0^{2 / 3} = 0$ , so $f$ is continuous at $x = 0$ .

Step 1: Set up the Difference Quotient at $x = 0$

\begin{aligned} \frac{f(0 + \Delta x) - f(0)}{\Delta x} &= \frac{(\Delta x)^{2/3} - 0}{\Delta x} \\ &= \frac{1}{(\Delta x)^{1/3}}. \end{aligned}
Step 2: Compute One-Sided Limits
For the left-hand derivative (

Δ x < 0

(Δ x)^{1 / 3}

is negative: f'_-(0) = \lim_{\Delta x\to 0^-} \frac{1}{(\Delta x)^{1/3}} = -\infty. For the right-hand derivative (

Δ x > 0

(Δ x)^{1 / 3}

is positive: f'_+(0) = \lim_{\Delta x\to 0^+} \frac{1}{(\Delta x)^{1/3}} = +\infty.

Step 3: Conclusion

Since f'_-(0) = -\infty and f'_+(0) = +\infty, the graph has infinite one-sided derivatives with opposite signs. The graph forms a vertical cusp pointing downwards at $(0, 0)$ , so f'(0) does not exist.

Geometric Graph:

The figure below illustrates the vertical cusp of $f (x) = x^{2 / 3}$ at $(0, 0)$ .

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Exercises

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