Question

Show that the values of the function $\frac{x^2+34x-71}{x^2+2x-7}$ do not lie between $5$ and $9$ for all, $x\in R$.

Answer 1

Let $\frac{x^2+34x-71}{x^2+2x-7}=y,x\in R$

$x^2+34x-71=y(x^2+2x-7)$

$x^2+34x-71=y{x}^2+2yx-7y$

$(y-1)x^2+(2y-34)x+71-7y=0$

As $x\in R$ discriminant must be greater than zero

${(2y-34)}^2\ge 4(y-1)(71-7y)$

${(y-17)}^2\ge (y-1)(71-7y)$

$y^2+289-34y\ge 71y-71-7y^2+7y$

${8y}^2-112y+360\ge 0$

$y^2-14y+45\ge 0$

$y^2-5y-9y+45\ge 0$

$y(y-5)-9(y-5)\ge 0$

$(y-5)(y-9)\ge 0$

So $y\in (-\infty ,5]\cup [9,\infty )$

So it can be in $(5,9)$