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Is this the right explanation for this question

1. Examine the continuity of the function


-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.

  • Calculus - ,

    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

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