Question

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{2} x+1, & x \leq 2 \\ 3-x, & x>2 \end{array}\right. $$

Short answer

The function \(f(x)\) is discontinuous at \(x = 2\). This discontinuity is not removable.

Step-by-step solution

Apply the x-value to the first part of the function

The first part of the function is defined for \(x \leq 2\). So, let's apply \(x=2\) to this part of the function: \(f(2)=\frac{1}{2} \cdot 2 +1=2\)

Apply the x-value to the second part of the function

The second part of the function is defined for \(x>2\). However, we still need to check what the value would be at \(x = 2\). So, let's apply \(x=2\) to this second part of the function (even though technically it's not within the defined domain for this part): \(f(2)=3-2=1\)

Compare the results

We can see that applying \(x = 2\) to both parts of the piecewise function did not result in the same value. Therefore, the function is discontinuous at \(x = 2\). As for the removability of this discontinuity, it is non-removable since we can't make the function continuous by defining or redefining it at a single point.