Can someone help me with this question. With steps.
Examine the continuity of the function
h(x)=
-x, if x<0
0, if x=0
x, if x>0
What would be discontinous at x=?
explain please
This is just h(x) = |x|
h is continuous at x=0, but
h' is discontinuous at x=0, since
h'(x) = -1 for x < 0
h'(x) = 1 for x > 0
Is there suppose to be calculations involved? :|
To examine the continuity of the function, we need to check the three conditions for continuity at x = 0:
1. Right-hand limit: We need to find the limit of the function as x approaches 0 from the right side (x > 0).
To do this, we substitute x = 0 into the function h(x) when x > 0:
h(0) = 0
Therefore, the limit as x approaches 0 from the right side is h(0) = 0.
2. Left-hand limit: We need to find the limit of the function as x approaches 0 from the left side (x < 0).
To do this, we substitute x = 0 into the function h(x) when x < 0:
h(0) = -0
Therefore, the limit as x approaches 0 from the left side is h(0) = 0.
3. Function value at x = 0: We need to find the actual value of the function at x = 0.
To do this, we substitute x = 0 into the function h(x):
h(0) = 0
Now, comparing the right-hand limit (0), left-hand limit (0), and the function value at x = 0 (0), we can see that they all match.
Since all of the conditions for continuity at x = 0 are met, the function h(x) is continuous at x = 0.