# Question f7666

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 $x \in \left(- \infty , \infty\right) \mathmr{and} y \in \left(- \infty , \infty\right)$

Let $y = \sqrt{x}$.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,

$f \left(x\right) \to 0$, as $x \to {0}_{+}$

The other side approach is unreal.

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

f(x) = 1/sqrt(x-2#), f is real, for $x > 2$. See the graph.
Also, as $x \to {2}_{+} , f \to \infty$.
So, f is continuous in the open interval $\left(2 , \infty\right)$