Find the intervals on which the function is continuous.
-$$y = \sqrt { x ^ { 2 } - 10 }$$
A) continuous on the interval $[ - \sqrt { 10 } , \sqrt { 10 } ]$
B) continuous on the intervals $( - \infty , - \sqrt { 10 } ]$ and $[ \sqrt { 10 } , \infty )$
C) continuous everywhere
D) continuous on the interval $[ \sqrt { 10 } , \infty )$

Question

Quizplus · Accepted Answer

The function is defined for $x\geq \sqrt{10}$. For the function to be continuous at a point $x_0$, three conditions must be satisfied:
1. $f(x_0)$ is defined
2. $\lim_{x	o x_0} f(x)$ exists
3. $\lim_{x	o x_0} f(x) = f(x_0)$

For the function $y=\sqrt{x^2-10}$, we have $f(x)$ defined for $x\geq \sqrt{10}$. Now we must check the limits as $x$ approaches $\sqrt{10}$ from both sides.

As $x	o \sqrt{10}^{-}$ (approaching $\sqrt{10}$ from the left), $y$ becomes undefined because we would be taking the square root of a negative number. Therefore the function is not continuous on $( - \infty , - \sqrt { 10 } ]$.

As $x	o \sqrt{10}^{+}$ (approaching $\sqrt{10}$ from the right), we have:
\begin{align*}
\lim_{x	o \sqrt{10}^{+}} \sqrt{x^2-10} &= \sqrt{\lim_{x	o \sqrt{10}^{+}} (x^2-10)}\
&= \sqrt{(\sqrt{10})^2-10}\
&= 0
\end{align*}

Therefore, the limit exists and $\lim_{x	o \sqrt{10}^{+}} \sqrt{x^2-10} = 0$. To check if the function is continuous at $x=\sqrt{10}$, we must check if $f(\sqrt{10})$ exists and if it is equal to the limit. We have $f(\sqrt{10}) = \sqrt{\sqrt{10}^2-10} = 0$, which is equal to the limit. Therefore, the function is continuous at $x=\sqrt{10}$ and on $[\sqrt{10}, \infty)$.

Thus, the function is continuous on the intervals $( - \infty , - \sqrt { 10 } ]$ and $[\sqrt { 10 } , \infty )$. Therefore, the best choice is 	extbf{(B)}.

Find the Intervals on Which the Function Is Continuous y=x2−10y = \sqrt { x ^ { 2 } - 10 }y=x2−10​

Find the Intervals on Which the Function Is Continuous $y = \sqrt { x ^ { 2 } - 10 }$