Question #f7666
1 Answer
Feb 1, 2017
c
Explanation:
Let y = f(x) = x, See the graph.
graph{x [-10, 10, -5, 5]} x
As seen, the graph is continuous,
with
Let
graph{sqrtx [-10, 10, -5, 5]}
You can see that y is real for x >=0.
We can reach O(0, 0), along the graph, only through positive values
Symbolically,
The other side approach is unreal.
So, the function is is seen as discontinuous. at x = 0.
In your example,
graph{(y-1/sqrt(x-2))(x-2-.001y)=0 [-10, 10, -5, 5]}
Also, as
f is seen as having infinite discontinuity at x = 2.
So, f is continuous in the open interval