# What is the breakpoint of a piecewise function?

For

$$f(x)=\left\{\begin{matrix} 0, & x\leq -1\\ \sqrt{1-x^2},& -1 < x < 1 \\ x, & x\geq 1 \end{matrix}\right.$$

My book says that the breakpoints are x = -1 and x = 1. How breakpoints are defined so that -1 and 1 is chosen?

• How does the book define "breakpoint"? Feb 14, 2019 at 15:49
• I would understand a "breakpoint" of a function $f(x)$ to be a location $x_b$ where the first derivative $f'(x_b)$ is disontinuous, i.e. where $f'(x)$ "jumps", mathematically, where the limits of $f'(x)$ for $x\to x_b$ from below is different from that from above. Feb 14, 2019 at 15:54
• OK, so it's the formula that has breakpoints. The parenthetical remark is introducing the terminology; in effect, it is the definition. Feb 14, 2019 at 15:55
• @Dr.WolfgangHintze, the formula $$f(x)=\begin{cases}0,\quad x\le0\\e^{-1/x}\quad x\gt0\end{cases}$$ has a breakpoint at $x=0$, even though the function has derivatives of all orders. Feb 14, 2019 at 15:58
• @Dr.WolfgangHintze, the passage that the OP quotes from their book talks about the formula having breakpoints, not the function. Feb 14, 2019 at 16:02

• @user247327 I worry that your definition would not include where $x=-1$ as both pieces of the function $=0$ there, so it would be continuous. Feb 14, 2019 at 16:18
• Don't worry! I said "or is not smooth". The derivative of f(x)= 0 is 0 while the derivative of $f(x)= \sqrt{1- x^2}$ is $-\frac{x}{\sqrt{1- x^2}}$ which does not exist at x= -1. Feb 14, 2019 at 17:13
Your function is defined piecewise. The break points are wherever one of the pieces ends and the next begins. Here, the first piece is defined for $$x\leq -1$$, so this piece ends and $$x=-1$$, and the next piece is defined for $$-1, so this piece ends at $$x=1$$.