Question #f7666

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Answer 1 · 2017-02-01T05:27:29

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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 $(2, oo)$