Quadratic inequality where $x\in\mathbb{R}$

Question

If $x$ belongs to the set of real numbers, show that $\dfrac x{x^2 - 5x +9}$ always lies in the interval $[\dfrac{-1}{11}, 1]$?

I tried to solve the problem using wavy curve method i.e by plotting the roots of the equation. We see that the denominator has a discriminant less than zero and hence it can be treated as a positive real number. When we treat the denominator as positive number and equate it to 0, we get x=0. I don't know what mistake I am making in my approach to the question?

The wavy curve method : https://i.stack.imgur.com/C2chw.jpg

Answer 1

I don't see a mistake in your solution. I think your solution is right:

Let $$\frac{x}{x^2-5x+9}=y.$$ Hence, the equation $$yx^2-(5y+1)x+9y=0$$ has real roots.

For $y=0$ we get $x=0$.

But, for $y\neq0$ we need $$(5y+1)^2-36y^2\geq0,$$ which gives $$-\frac{1}{11}\leq y\leq1.$$

Also, we can use the following idea.

For $x>0$ by AM-GM we obtain: $$\frac{x}{x^2-5x+9}=\frac{1}{x+\frac{9}{x}-5}\leq\frac{1}{2\sqrt{x\cdot\frac{9}{x}}-5}=1.$$ For $x<0$ by AM-GM again we obtain: $$\frac{x}{x^2-5x+9}=\frac{1}{x+\frac{9}{x}-5}\geq\frac{1}{-2\sqrt{(-x)\left(-\frac{9}{x}\right)}-5}=-\frac{1}{11}.$$ For $x=0$ we get a value $0$ and since $f(x)=\frac{x}{x^2-5x+9}$ is a continuous function, we are done.

Answer 2

Note that the method of intervals (wavy curve method) is used to solve an inequality. The solution of the inequality is indeed: $$\dfrac x{x^2 - 5x +9}>0 \ \ (<0) \Rightarrow x>0 \ \ (x<0).$$ However, the problem is asking to show the range: $$-\frac1{11}\le \dfrac x{x^2 - 5x +9}\le 1.$$ You can find the minimum and maximum values of the function: $f(x)=\dfrac x{x^2 - 5x +9}$ by using derivative: $$\begin{align}f'(x)&=\frac{x^2-5x+9-x(2x-5)}{(x^2-5x+9)^2}=\frac{9-x^2}{(x^2-5x+9)^2}=0 \Rightarrow x=\pm3;\\ f''(-3)&=\frac{2}{363}>0 \Rightarrow f(-3)=-\frac{1}{11} \ (min);\\ f''(3)&=-\frac{2}{3}<0 \Rightarrow f(3)=1 \ (max). \end{align}$$