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?
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1$\begingroup$ How does the book define "breakpoint"? $\endgroup$– Barry CipraFeb 14, 2019 at 15:49
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1$\begingroup$ 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. $\endgroup$– Dr. Wolfgang HintzeFeb 14, 2019 at 15:54
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1$\begingroup$ OK, so it's the formula that has breakpoints. The parenthetical remark is introducing the terminology; in effect, it is the definition. $\endgroup$– Barry CipraFeb 14, 2019 at 15:55
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1$\begingroup$ @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. $\endgroup$– Barry CipraFeb 14, 2019 at 15:58
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1$\begingroup$ @Dr.WolfgangHintze, the passage that the OP quotes from their book talks about the formula having breakpoints, not the function. $\endgroup$– Barry CipraFeb 14, 2019 at 16:02
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2 Answers
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"Break points" are where the graph "breaks"- where the graph is no longer continuous or is not "smooth". Basically that means where the "formula" for the function changes so that you need a "piecewise" definition.
answered Feb 14, 2019 at 15:50
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$\begingroup$ But how break point is mathematically defined? How can I get the breakpoints of a given function if there is any? $\endgroup$ Feb 14, 2019 at 15:57
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$\begingroup$ @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. $\endgroup$ Feb 14, 2019 at 16:18
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$\begingroup$ 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. $\endgroup$ Feb 14, 2019 at 17:13
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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 $-1<x<1$, so this piece ends at $x=1$.
You could then say that the break points are where the function is broken into its constituent parts.
answered Feb 14, 2019 at 16:16