Asked by Matthew
Is this the right explanation for this question
1. Examine the continuity of the function
h(x)=
-x, if x<0
0, if x=0
x, if x>0
Answer: Since h(x)=
-x, if x<0
0, if x=0
x, if x>0 is h=|x|,
we can conclude that the function is
1. continuous everywhere
2. discontinuous at x=0 because limit from the left (-1) and limit from the right (+1) does not equal the same value, therefore it doesn't exist.
1. Examine the continuity of the function
h(x)=
-x, if x<0
0, if x=0
x, if x>0
Answer: Since h(x)=
-x, if x<0
0, if x=0
x, if x>0 is h=|x|,
we can conclude that the function is
1. continuous everywhere
2. discontinuous at x=0 because limit from the left (-1) and limit from the right (+1) does not equal the same value, therefore it doesn't exist.
Answers
Answered by
Steve
h is continuous everywhere
h' is discontinuous at x=0
the limit of h(x) as x->0 is the same (0) from both sides, and is defined as 0 at x=0, so h is continuous.
h' = -1 for x<0
h' = 1 for x > 0
but h(x) is continuous
h' is discontinuous at x=0
the limit of h(x) as x->0 is the same (0) from both sides, and is defined as 0 at x=0, so h is continuous.
h' = -1 for x<0
h' = 1 for x > 0
but h(x) is continuous
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