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Define $g(x)$ as follows: \[ g(x)= \begin{cases} -1, & \text{if } x < 0,\\ 0, & \text{if } x = 0, \\ 1, & \text{if } x > 0. \\ \end{cases} \]Let $f(x) = g(x+1) - g(x-1).$ Compute $f\left(\frac12\right) + f\left(-\frac12\right).$

 Sep 12, 2022
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\(f(x)\ =\ g(x+1)-g(x-1)\\~\\ \color{purple}{f(\frac12)}\ =\ g(\frac12+1)-g(\frac12-1)\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ g(\frac32)-g(-\frac12)\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ 1--1\\~\\ \color{purple}\phantom{f(\frac12)}\ =\ 2\\~\\ \color{Green}{f(-\frac12)}\ =\ g(-\frac12+1)-g(-\frac12-1)\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ g(\frac12)-g(-\frac32)\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ 1--1\\~\\ \color{Green}\phantom{f(\frac-12)}\ =\ 2\\~\\ f(\frac12)+f(-\frac12)\ =\ 2+2\ =\ 4\)

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 Sep 13, 2022

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