Question

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

Answer 1

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

Explanation:

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}`