Question #f7666

1 Answer
Feb 1, 2017

Answer:

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)#