Question

activity: indicate the domain of the given function by marking the appropriate interval or intervals on the number line.
$f(x)=x + \frac{1}{x}$
$f(x)=\frac{x^{2}}{x^{2}+1}$

Explanation:

Step1: Analyze first - function

For the function $f(x)=x + \frac{1}{x}$, the denominator of the fraction cannot be zero. Since the denominator is $x$, $x
eq0$.

Step2: Analyze second - function

For the function $f(x)=\frac{x^{2}}{x^{2}+1}$, the denominator $x^{2}+1$ is always positive. Because $x^{2}\geq0$ for all real $x$, then $x^{2}+1\geq1$ for all real $x$. So $x$ can be any real number.

Answer:

For $f(x)=x+\frac{1}{x}$, the domain is $(-\infty,0)\cup(0,\infty)$. Mark the number - line with an open circle at $x = 0$ and shade the regions to the left and right of $x = 0$.
For $f(x)=\frac{x^{2}}{x^{2}+1}$, the domain is $(-\infty,\infty)$. Mark the entire number - line.