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Where is the graph of [tex]f(x)=4\lfloor x-3\rfloor+2[/tex] discontinuous?

A. All real numbers
B. All integers
C. Only at multiples of 3
D. Only at multiples of 4

Sagot :

GinnyAnsweridn
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.
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