Answer
$y=\frac{1}{|x|+1}-\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$
Work Step by Step
$$y=\frac{1}{|x|+1}-\frac{x^2}{2}$$
- Domain: because $|x|\ge0$, $|x|+1\gt0$, so the function is defined on $(-\infty,\infty)$
1) Examine $f(x)=\frac{1}{|x|+1}$:
As proved above, $|x|+1\ne0$ for all $x\in(-\infty,\infty)$.
Therefore, $\lim_{x\to c}\frac{1}{|x|+1}=\frac{1}{|c|+1}$ for all $x\in(-\infty,\infty)$
So the function $f(x)=\frac{1}{|x|+1}$ is continuous on $(-\infty,\infty)$.
2) Examine $g(x)=\frac{x^2}{2}$
$\lim_{x\to c}\frac{x^2}{2}=\frac{c^2}{2}$ for all $x\in(-\infty,\infty)$
So the function $g(x)=\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$.
According to Theorem 8, the subtraction of two continous function at $x=c$ is also continuous at $x=c$.
Therefore, $y=\frac{1}{|x|+1}-\frac{x^2}{2}$ is continuous on $(-\infty,\infty)$
