The range of the function $f(x) = \frac{x^2}{x^2+1}$ is

Let $f: R - \{2, 6\} \rightarrow R$ be a real-valued function defined as $f(x) = \frac{x^2+2x+1}{x^2-8x+12}$. Then the range of $f$ is

Define the real-valued function $f: R - \{0\} \rightarrow R$ defined by $f(x) = \frac{1}{x}$,where $x \in R - \{0\}$. Complete the table given below using this definition. What is the domain and range of this function?

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	....	....	....	....	....	....	....	....	....

Let $f : R \to R$ be defined by $f(x) = \frac{x}{1 + x^2}, x \in R$. Then the range of $f$ is

Let $f(x) = \cos(\pi(|x| + 2[x]))$,where $[.]$ represents the greatest integer function. Then:

The set of all real values of $x$ for which the real-valued function $f(x) = \left(1 + \frac{1}{x}\right)^x$ is defined,is