Determine the value of [tex]$x$[/tex] for which the function [tex]$f(x)=\left\{\begin{array}{l}5 x, \text { if } x<0 \\ 1, \text { if } x=0 \\ -5 x, \text { if } x>0\end{array}\right.$[/tex] is discontinuous.
a) 1
b) 0
c) 5
d) -5
Answer (2)
Determine the left-hand limit as x approaches 0: lim x → 0 − f ( x ) = 0 .
Determine the right-hand limit as x approaches 0: lim x → 0 + f ( x ) = 0 .
Evaluate the function at x = 0 : f ( 0 ) = 1 .
Since lim x → 0 f ( x ) = f ( 0 ) , the function is discontinuous at x = 0 , so the answer is 0 .
Explanation
Analyze the function We are given a piecewise function:
0\end{array}\right."> f ( x ) = ⎩ ⎨ ⎧ 5 x , if x < 0 1 , if x = 0 − 5 x , if x > 0
We need to determine the value of x for which this function is discontinuous.
Check continuity at x=0 A function is discontinuous at a point if the limit of the function as x approaches that point does not exist, or if the limit exists but is not equal to the function's value at that point. In this case, we need to check the continuity at x = 0 since the function is defined differently for x < 0 , x = 0 , and 0"> x > 0 .
Calculate limits First, let's find the left-hand limit as x approaches 0:
lim x → 0 − f ( x ) = lim x → 0 − 5 x = 5 ( 0 ) = 0
Next, let's find the right-hand limit as x approaches 0:
lim x → 0 + f ( x ) = lim x → 0 + − 5 x = − 5 ( 0 ) = 0
Since the left-hand limit and the right-hand limit are both equal to 0, the limit exists and is equal to 0:
lim x → 0 f ( x ) = 0
Compare limit and function value Now, let's check the value of the function at x = 0 :
f ( 0 ) = 1
Since the limit as x approaches 0 is 0, but the function's value at x = 0 is 1, the function is discontinuous at x = 0 .
Conclusion For x < 0 , f ( x ) = 5 x is a linear function, and linear functions are continuous everywhere. Similarly, for 0"> x > 0 , f ( x ) = − 5 x is a linear function and is continuous everywhere.
Therefore, the only point of discontinuity is at x = 0 .
Examples
In signal processing, discontinuous functions can model abrupt changes in signals. For example, the Heaviside step function, which is 0 for negative time and 1 for positive time, is used to represent a sudden switch being turned on. Understanding the points of discontinuity is crucial for analyzing the behavior of such signals and designing appropriate filters or control systems.
The function is discontinuous at x = 0 because the limit as x approaches 0 does not equal the function value at that point. Thus, the answer is 0 .
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