Question

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

Answer 1

`f(A) = (1,+oo)`

Explanation:

`f(x)=(x^2+1)/x^2` , `A=(-oo,0)uu(0,+oo)`

`f'(x)=((x^2+1)'x^2-(x^2)'(x^2+1))/x^4=`

`(2x^3-2x^3-2x)/x^4=`

`-2/x^3`

For `x>0` we have `f'(x)<0` so `f` is strictly decreasing in `(0,+oo)`

For `x<0` we have `f'(x)>0` so `f` is strictly increasing in `(-oo,0)`

`A_1=(-oo,0)`, `A_2=(0,+oo)`

`lim_(xrarr0^(-))f(x)=lim_(xrarr0^(-))(x^2+1)/x^2=+oo`
`lim_(xrarr0^(+))f(x)=lim_(xrarr0^(+))(x^2+1)/x^2=+oo`
`lim_(xrarr-oo)f(x)=lim_(xrarr-oo)(x^2+1)/x^2=lim_(xrarr-oo)x^2/x^2=1`
`lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(x^2+1)/x^2=1`

`f(A_1)=f(((-oo,0)))=(lim_(xrarr-oo)f(x),lim_(xrarr0^(-))f(x))=`

`(1,+oo)`

`f(A_2)=f(((0,+oo)))=(lim_(xrarr+oo)f(x),lim_(xrarr0^+)f(x))=(1,+oo)`

Range `=f(A)=f(A_1)uuf(A_2)=(1,+oo)`