Find the range of `f(x)=(x^2+34 x-71)/(x^3+2x-7)`

Answer

`(-oo, 5] cup [9,oo)`

Explanation

Let `(x^(2)+34x-71)/(x^(2)+2x-7)=y`
or ` x^(2)(1-y)+2(17-y)x+(7y-71)=0`
For the real value of x,
`b^(2)-4ac ge0`
or `4(17-y)^(2)-4(1-y)(7y-71) ge 0`
or ` y^(2)-14y+45 ge 0`
i.e., `y le 5 " or " y ge 9`
Hence, range is `(-oo,5] cup [9,oo).`