Question

Question #f7666

Answer 1

c

Explanation:

Let y = f(x) = x, See the graph.

graph{x [-10, 10, -5, 5]} x

As seen, the graph is continuous,

with `x in (-oo, oo) and y in (-oo, oo)`

Let `y =sqrtx`.See the graph.

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,

`f(x) to 0`, as `x to 0_+`

The other side approach is unreal.

So, the function is is seen as discontinuous. at x = 0.

In your example,

`f(x) = 1/sqrt(x-2`), f is real, for `x > 2`. See the graph.

graph{(y-1/sqrt(x-2))(x-2-.001y)=0 [-10, 10, -5, 5]}

Also, as `x to 2_+ , f to oo`.

f is seen as having infinite discontinuity at x = 2.

So, f is continuous in the open interval `(2, oo)`