Incorrect ways to determine if a piecewise function is differentiable at a point?

Given a piecewise function, let's say:

$$f(x)= \begin{cases} x^2 \quad x < 0 \\ x^3 \quad x \ge 0 \end{cases}$$

Let's say we know the function is continuous and we only deal with real numbers.

I have seen people check if it's differentiable at 0 in two different ways, but I'm not sure if all are correct. So help me out figuring out which are actually correct ways to do it.

Method 1: Use the definition of the derivative, one for x>0 and one for x<0. Check if the the limits you get are the same.

Method 2: Simply differentiate the function like this:

$$f(x)= \begin{cases} 2x\quad x < 0 \\ 3x^2 \quad x > 0 \end{cases}$$

Then check if it's continuous in 0 by checking the limit from both sides.

Which methods are correct? And are there other (analytic) methods?

• @Rahul: Actually, the methods are not equivalent as method $(2)$ checks if the function is continuously differentiable at $x = 0$ while method $(1)$ checks only if the function is differentiable at $x = 0$. Apr 27 '17 at 18:04
• @levap: You're right. I misunderstood the statement "one for x>0 and one for x<0"; I would find it clearer if the limiting variable was called $h$ or $\Delta x$ which is more standard.
– user856
Apr 27 '17 at 21:09

Method $(1)$ is correct because it uses the definition of differentiability as the existence of a certain limit and check its existence by checking the existence and agreement of two one-sided limits.

Method $(2)$ works under certain assumptions on $f$ which are most often met. However, it only gives you a sufficient criterion for the differentiability of $f$, not a necessary one. Assume the problematic point is $x_0$ and that the function $f$ is continuous on some interval containing $x_0$ and differentiable on some punctured interval containing $x_0$. Method $(2)$ suggests that if

$$\lim_{x \to x_0^{-}} f'(x) = \lim_{x \to x_0^{+}} f'(x)$$

then $f$ is differentiable at $x = x_0$ and

$$f'(x_0) = \lim_{x \to x_0^{-}} f'(x) = \lim_{x \to x_0^{+}} f'(x).$$

This is indeed true. Let $h > 0$ be small enough. Since $f$ is continuous on $[x_0, x_0 + h]$ and differentiable on $(x_0,x_0 + h)$, the mean value theorem implies that

$$\frac{f(x_0 + h) - f(x_0)}{h} = f'(c_h)$$

for some $c_h \in (x_0, x_0 + h)$. As $h \to 0^{+}$, we see that $c_h \to x_0$ so if the limit $\lim_{x \to x_0^{+}} f'(x)$ exists we have

$$f'_{+}(x_0) = \lim_{h \to 0^{+}} \frac{f(x_0 + h) - f(x_0)}{h} = \lim_{h \to 0^{+}} f'(c_h) = \lim_{x \to x_0^{+}} f'(x_0).$$

Repeating the same argument for $f'_{-}(x_0)$ we have proved that method $(2)$ works under the assumptions above.

However, it might be that one of the limits $\lim_{x \to x_0^{-}} f'(x), \lim_{x \to x_0^{+}} f'(x)$ fails to exist and yet the function $f$ is differentiable at $x = x_0$. The reason is that method $(2)$ actually checks if $f$ is continuously differentiable at $x = x_0$, not only differentiable! For example, if

$$f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & x \neq 0, \\ 0 & x = 0 \end{cases}$$

then $f$ is differentiable at $x = 0$ with $f'(0) = 0$ but

$$\lim_{x \to 0^{\pm}} f'(x) = \lim_{x \to 0^{\pm}} \left( 2x \sin \left( \frac{1}{x} \right) - \sin \left( \frac{1}{x} \right) \right)$$

do not exist and $f$ is not continuously differentiable at $x = 0$.

• Very nice explanation, I appreciate the examples, thank you. May 3 '17 at 9:27