at what points is the given function continuous?\ny = \\frac{\\sqrt{x^{4}+4}}{1 + \\sin^{2}x}\nthe set of x…

Explanation:

Step1: Recall continuity conditions

A rational - function $\frac{f(x)}{g(x)}$ is continuous where $g(x)\neq0$. Here, $f(x)=\sqrt{x^{4}+4}$ and $g(x)=1 + \sin^{2}x$.

Step2: Analyze the denominator

We know that $- 1\leqslant\sin x\leqslant1$, so $0\leqslant\sin^{2}x\leqslant1$. Then, $1+\sin^{2}x\geqslant1$ for all real - valued $x$. Since $1+\sin^{2}x$ is never zero for any real number $x$.

Step3: Analyze the numerator

The function $y = \sqrt{x^{4}+4}$ is defined for all real $x$ because $x^{4}\geqslant0$ for all real $x$, and $x^{4}+4\geqslant4>0$ for all real $x$.

Answer:

$(-\infty,\infty)$