To find where the function [tex]\( f(x) = 4\lfloor x-3 \rfloor + 2 \)[/tex] is discontinuous, we need to consider where the floor function, [tex]\( \lfloor x-3 \rfloor \)[/tex], changes its value.
The floor function, [tex]\( \lfloor x \rfloor \)[/tex], returns the greatest integer less than or equal to [tex]\( x \)[/tex]. For any given integer [tex]\( n \)[/tex], [tex]\( \lfloor x \rfloor \)[/tex] is discontinuous at [tex]\( x = n \)[/tex].
Let's rewrite the floor function in the given equation: [tex]\[ f(x) = 4\lfloor x-3 \rfloor + 2 \][/tex]
The floor function [tex]\( \lfloor x-3 \rfloor \)[/tex] changes its value when [tex]\( x-3 \)[/tex] is an integer. We can find these points by setting [tex]\( x-3 = n \)[/tex], where [tex]\( n \)[/tex] is any integer. Solving for [tex]\( x \)[/tex], we get: [tex]\[ x = n + 3 \][/tex]
Hence, the function [tex]\( f(x) = 4\lfloor x-3 \rfloor + 2 \)[/tex] is discontinuous for every integer value of [tex]\( x \)[/tex].
Answer: The function [tex]\( f(x) \)[/tex] is discontinuous at all integers.
