How do you find the range of $f (x) = \frac{1}{x^{2} + 9}$ $f(x)= 1/(x^2 + 9)$ ?

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Answer 1 · 2015-09-16T15:49:23

Range ${ y in RR$ , 0< y<= 1/9}#

Solve $y= 1/(x^2+9)$ for x

$x^2= (1-9y)/y$ ,

$x= sqrt((1-9y)/y)$

This implies that $y !=0$ and $1-9y>=0$ , that is $y <= 1/9$ Hence range of f(x) would be

${ y in RR,0< y <= 1/9}$

How do you find the range of f(x)= 1/(x^2 + 9)f(x)=1x2+9f(x)= 1/(x^2 + 9)?