Matching Pieces of a Piecewise Function for $C^n$ Smoothness

A piecewise function

$g(x) = \begin{cases} L(x), & x \le a, \\ R(x), & x > a, \end{cases}$

is built from two smooth formulas $L$ and $R$ that meet at the breakpoint $x = a$. Away from $a$ each piece is as smooth as you like, so the only place smoothness can fail is at the seam. The question is always: how many matching conditions at $a$ make $g$ belong to $C^{n}$ (i.e. have a continuous $n$-th derivative)?

The matching principle

$g \in C^{n}$ at $x = a$ $\iff$ the two pieces agree at $a$ in value and in every derivative up to order $n$: $L^{(k)}(a) = R^{(k)}(a) \quad\text{for } k = 0, 1, 2, \dots, n.$

That is $n+1$ equations (orders $0$ through $n$). The logic chains downward: a continuous $g^{(n)}$ forces $g^{(n-1)}, \dots, g'$ and $g$ to be continuous too, because a function with a derivative is automatically continuous. So you never check continuity of the lower derivatives separately — matching the top order plus all the ones beneath it is exactly the full list.

Intuitively, at a breakpoint each piece supplies a one-sided value $L^{(k)}(a^{-})$ from the left and $R^{(k)}(a^{+})$ from the right. The derivative $g^{(k)}$ exists and is continuous at $a$ only when these one-sided values coincide for that order and all lower ones.

How to apply it

1. Identify the breakpoint $a$ and the two pieces $L$ (used for $x \le a$) and $R$ (used for $x > a$).
2. Decide the target order $n$: "continuous" is $C^{0}$ (match $k=0$); "differentiable" / "continuous first derivative" is $C^{1}$ (match $k=0,1$); "continuous second derivative" is $C^{2}$ (match $k=0,1,2$); and so on.
3. Differentiate each piece up to order $n$ and evaluate at $a$. Watch the product, chain, and quotient rules — most slips happen here.
4. Set $L^{(k)}(a) = R^{(k)}(a)$ for $k = 0, \dots, n$, giving $n+1$ equations, then solve for the unknown constants.

Worked mini-example

Choose $b$ and $c$ so that

$g(x) = \begin{cases} \sin x, & x \le 0, \\ b x + c, & x > 0, \end{cases}$

is $C^{1}$ (continuously differentiable). Here $n = 1$, so we match orders $k = 0$ and $k = 1$ at $a = 0$.

Left piece $L(x) = \sin x$: $\;L(0) = 0,\; L'(x) = \cos x,\; L'(0) = 1.$

Right piece $R(x) = b x + c$: $\;R(0) = c,\; R'(x) = b,\; R'(0) = b.$

Matching:

So $g$ is $C^{1}$ exactly when $b = 1,\, c = 0$, i.e. the right piece is the tangent line $y = x$. (For a $C^{2}$ glue we would additionally need $R''(0) = L''(0)$, i.e. $0 = -\sin 0 = 0$, which already holds — but a generic quadratic right piece would pin a third condition.)

This is the same machinery the konkur problem uses: there the left piece is $f(x)e^{x}$ and the right is $3x^{2}-4x+7$, and demanding a continuous second derivative ($C^{2}$) gives the three equations $L(0)=7,\ L'(0)=-4,\ L''(0)=6$ from the product rule.

Exam trick: Translate the wording into a number first — "$g \in C^{n}$" means "match orders $0$ through $n$," giving $n+1$ equations, no more, no less. The biggest trap is matching only the top order (e.g. just $g''$) and forgetting that you still need the value and first-derivative equations; conversely, do not waste time re-checking continuity of $g$ once you have matched a higher derivative, since that comes for free.