Question

If $x$ is real, prove that $\frac{x}{x^2-5x+9}$ lies between $-\frac{1}{11}$ and $1$.

Answer 1

let $\frac{x}{x^2-5x+9}=y$

$x=y{x}^2-\left(5y\right)x+9y$

$y{x}^2-(1+5y)x+9y$

For this quadratic equation to have real roots

Discriminant must be $\ge 0$

${(-1-5y)}^2-4\times y\times 9y\ge 0$

$1+25y^2+10y-36y^2\ge 0$

$1+10y-11y^2\ge 0$

$1-y+11y-11y^2\ge 0$

$(1-y)+11y(1-y)\ge 0$

$(11y+1)(y-1)\le 0$

So $(y-(\frac{-1}{11}\left)\right)(y-1)\le 0$

So $y\in [\frac{-1}{11},1]$