At which value of \(x\) does the graph of \(f(x)=\operatorname{sgn}(2x-8)\) have a jump discontinuity?

Correct answer: \(x=4\)

The signum function \(\operatorname{sgn}(u)\) equals \(-1\) for \(u<0\) and \(1\) for \(u>0\), so its jump occurs where \(u=0\). Here, \(2x-8=0\), giving \(x=4\). At \(x=4\), the function value is \(0\), while the left- and right-hand limits are \(-1\) and \(1\), respectively. Exam tip: To locate the jump of a signum function, set its inner expression equal to zero.