# Find the range of the function f(x) = (1+ x^2)/x^2 ?

May 10, 2018

$f \left(A\right) = \left(1 , + \infty\right)$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 1}{x} ^ 2$ , $A = \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

$f ' \left(x\right) = \frac{\left({x}^{2} + 1\right) ' {x}^{2} - \left({x}^{2}\right) ' \left({x}^{2} + 1\right)}{x} ^ 4 =$

$\frac{2 {x}^{3} - 2 {x}^{3} - 2 x}{x} ^ 4 =$

$- \frac{2}{x} ^ 3$

For $x > 0$ we have $f ' \left(x\right) < 0$ so $f$ is strictly decreasing in $\left(0 , + \infty\right)$

For $x < 0$ we have $f ' \left(x\right) > 0$ so $f$ is strictly increasing in $\left(- \infty , 0\right)$

${A}_{1} = \left(- \infty , 0\right)$, ${A}_{2} = \left(0 , + \infty\right)$

${\lim}_{x \rightarrow {0}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {0}^{-}} \frac{{x}^{2} + 1}{x} ^ 2 = + \infty$
${\lim}_{x \rightarrow {0}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {0}^{+}} \frac{{x}^{2} + 1}{x} ^ 2 = + \infty$
${\lim}_{x \rightarrow - \infty} f \left(x\right) = {\lim}_{x \rightarrow - \infty} \frac{{x}^{2} + 1}{x} ^ 2 = {\lim}_{x \rightarrow - \infty} {x}^{2} / {x}^{2} = 1$
${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \frac{{x}^{2} + 1}{x} ^ 2 = 1$

$f \left({A}_{1}\right) = f \left(\left(\left(- \infty , 0\right)\right)\right) = \left({\lim}_{x \rightarrow - \infty} f \left(x\right) , {\lim}_{x \rightarrow {0}^{-}} f \left(x\right)\right) =$

$\left(1 , + \infty\right)$

$f \left({A}_{2}\right) = f \left(\left(\left(0 , + \infty\right)\right)\right) = \left({\lim}_{x \rightarrow + \infty} f \left(x\right) , {\lim}_{x \rightarrow {0}^{+}} f \left(x\right)\right) = \left(1 , + \infty\right)$

Range $= f \left(A\right) = f \left({A}_{1}\right) \cup f \left({A}_{2}\right) = \left(1 , + \infty\right)$