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

Set $y=\dfrac{x}{x^2-5x+9}$, clear denominators to get a quadratic in $x$, and require its discriminant $\ge0$ (for real $x$) — this bounds $y$.

Note $x^2-5x+9$ has discriminant $25-36=-11<0$, so it is never zero for real $x$ (always positive, since the leading coefficient is positive) — the given expression is defined for every real $x$.

Let $y=\dfrac{x}{x^2-5x+9}$. Then

$y(x^2-5x+9)=x \;\Rightarrow\; yx^2-(5y+1)x+9y=0.$

For this to have a real solution $x$ (which it must, since $y$ arises from a real $x$), the discriminant must be $\ge0$ when $y\ne0$:

$(5y+1)^2-4(y)(9y)\ge0 \;\Rightarrow\; 25y^2+10y+1-36y^2\ge0 \;\Rightarrow\; -11y^2+10y+1\ge0,$

i.e. $11y^2-10y-1\le0$. Solving $11y^2-10y-1=0$:

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