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.

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

The given explanation for the continuity of the function is correct. However, let me explain how we can determine the continuity of the function using the concept of limits.

To determine the continuity of a function, we need to check three conditions:

1. The function should be defined at the point in question.
2. The left-hand limit and the right-hand limit at that point should exist.
3. The left-hand limit, right-hand limit, and the value of the function at that point should be equal.

Let's apply these conditions to the function h(x) = -x, if x < 0, 0, if x = 0, x, if x > 0.

1. The function is defined for all values of x. So, it is defined at x = 0.
2. To find the left-hand limit as x approaches 0, we consider the values of x that are smaller than 0. In this case, h(x) = -x. So, as x approaches 0 from the left (negative values of x), the function approaches 0. Therefore, the left-hand limit is 0.
To find the right-hand limit as x approaches 0, we consider the values of x that are greater than 0. In this case, h(x) = x. So, as x approaches 0 from the right (positive values of x), the function also approaches 0. Therefore, the right-hand limit is 0.
3. The value of the function at x = 0 is defined as 0.

Now, applying the three conditions:
- The function is defined at x = 0.
- The left-hand limit as x approaches 0 is 0.
- The right-hand limit as x approaches 0 is 0.

Since all three conditions are satisfied, we can conclude that the function h(x) is continuous at x = 0.

For any other value of x (x < 0 or x > 0), the function is simply the absolute value of x (h(x) = |x|), which is continuous everywhere.

Therefore, the correct answer is that the function h(x) is continuous everywhere, including at x = 0.