Question

In Exercises \(19-24,(\) a) find each point of discontinuity. (b) Which of the discontinuities are removable? not removable? Give reasons for your answers. $$f(x)=\left\\{\begin{array}{ll}{3-x,} & {x<2} \\ {\frac{x}{2}+1,} & {x>2}\end{array}\right.$$

Short answer

The function \( f(x) \) is discontinuous at \( x = 2 \) and the discontinuity is not removable because the left-hand limit does not equal the right-hand limit.

Step-by-step solution

Identify the Domains

Firstly, the domain of the function need to be identified. For the given function it is all real numbers, as there are no restrictions placed on \( x \).

Find the limit at x=2

Next, the limit of the function as \( x \) approaches 2 should be determined. Since the function is defined differently for values less than 2 and values greater than 2, both the left-hand limit and the right-hand limit need to be calculated. The left-hand limit is the limit of the function as \( x \) approaches 2 from the left, which can be calculated as \( \lim_{{x \to 2^-}} (3 - x) = 1 \). The right-hand limit is the limit of the function as \( x \) approaches 2 from the right, which can be calculated as \( \lim_{{x \to 2^+}} (\frac{x}{2} + 1) =2 \). The function is discontinuous at \( x = 2 \), because the left-hand limit does not equal the right-hand limit.

Determine the nature of the discontinuity

Finally, determine whether the discontinuity at \( x = 2 \) is removable or not. A discontinuity is removable if the function value at that point can be redefined to make the function continuous. In this case, there is no value for \( f(2) \) that would make the function continuous at \( x = 2 \), because the left-hand limit does not equal the right-hand limit. Therefore, the discontinuity is not removable.