Continuity at the junction points

On the open pieces $I_k$ the function inherits the continuity of the formula $f_k$. At the junction points between two neighbouring pieces you must check by hand that the two values coincide: if at the boundary point $x_0$ between $I_k$ and $I_{k+1}$ we have

$\lim_{x\to x_0^-}f_k(x) = f(x_0) = \lim_{x\to x_0^+}f_{k+1}(x),$

the function is continuous at $x_0$; otherwise it has a jump.

Example — Finding a parameter to obtain continuity

$f(x)=\begin{cases} 2x+a & x\le 1 \\ x^2 & x>1\end{cases}$. For which $a$ is the function continuous at $x=1$?

On the left $\lim_{x\to 1^-}(2x+a)=2+a$; on the right $\lim_{x\to 1^+}x^2=1$; continuity $\iff 2+a=1\iff \boxed{a=-1}$.

Example — Intersection with a line

Find the points of $f(x)=\begin{cases} -x+1 & x<0 \\ x^2 & x\ge 0\end{cases}$ that satisfy $y=1$.

On $x<0$: $-x+1=1 \iff x=0$, outside the interval $\to$ discarded. On $x\ge 0$: $x^2=1 \iff x=1$ ($x=-1$ outside). Point $(1,1)$.

Moreover the piece $x<0$ has a right-hand limit at $0$: $\lim_{x\to 0^-}(-x+1)=1$ — the value $1$ is “approached” from the left without being reached. It is important to read the domain of the piece.

Links

Topics: Functions and properties
Concepts: Continuity · Piecewise function
Skills: Reasoning by cases · Solving equations