find the intervals on which the function is continuous.\ny = \\frac{2}{x^{2}-16}\ndiscontinuous only when x…

Explanation:

Step1: Identify denominator condition

A rational - function $y = \frac{f(x)}{g(x)}$ is discontinuous when $g(x)=0$. Here, $y=\frac{2}{x^{2}-16}$, and $g(x)=x^{2}-16$.

Step2: Solve for $x$ when $g(x) = 0$

Set $x^{2}-16 = 0$. Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we have $(x + 4)(x - 4)=0$. Then, by the zero - product property, $x+4 = 0$ or $x - 4=0$, so $x=-4$ or $x = 4$.

Answer:

discontinuous only when $x=-4$ or $x = 4$