Question

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\left\\{\begin{array}{ll} \csc \frac{\pi x}{6}, & |x-3| \leq 2 \\ 2, & |x-3|>2 \end{array}\right. $$

Short answer

The function \(f(x)\) is not continuous at \(x = 3\), and this discontinuity is not removable.

Step-by-step solution

Analyze the Discontinuities of the Cosecant Function

Let's first examine the function \(\csc(\frac{\pi x}{6})\). This is f(x) when \( |x-3| \leq 2 \), which corresponds to the interval \( 1 \leq x \leq 5 \). The cosecant function has discontinuities whenever its argument is a multiple of \(\pi \). Thus, the function \(\csc(\frac{\pi x}{6})\) will be not continuous whenever \(\frac{\pi x}{6}\) is a multiple of \(\pi \), specifically when \(x\) is an integer multiple of 6. However, between \( 1 \leq x \leq 5 \), there are no multiple of 6. Therefore, within this interval, there are no inherent discontinuities due to the cosecant function.

Make Sure the Constant Function is Always Continuous

The function 2 is f(x) when \( |x-3| > 2 \), which corresponds to \( x < 1 \) and \( x > 5 \). This function is always continuous as it is a constant function.

Check Continuity at Boundary points

The boundary point is at \( x = 3 \). At this point, the function is defined to be \(\csc(\frac{\pi x}{6})\), which gives \(\csc(\frac{\pi(3)}{6}) = \csc(\frac{\pi}{2}) = 1\). However, as we approach \(x = 3\) from values larger than 3 (from the second case), the function approaches 2, not 1. Therefore, \(x = 3\) is a discontinuity, and it is not removable as the function does not approach the same value from both directions.