Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$

Question

Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$

I have reached: $x^2(y-1) + x(y-3) + (y-1) = 0$

I also know that the denominator of $y= f(x)$ is greater than $0$.

How do I continue from here? I am unable to find a suitable condition to proceed.

Edit: vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$

and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$ but that ain't much helpful, is it?

Answer 1

You have shown that $y$ is in the range of the function
$$y = \frac{x^2 + 3x + 1}{x + 1}$$ if $x$ is a real-valued root of the quadratic equation $$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$ For $x$ to be a real-valued root of the quadratic equation, its discriminant must be nonnegative. The discriminant is \begin{align*} \Delta & = b^2 - 4ac\\ & = (y - 3)^2 - 4(y - 1)(y - 1)\\ & = y^2 - 6y + 9 - 4(y^2 - 2y + 1)\\ & = -3y^2 + 2y + 5 \end{align*} Hence, we require that \begin{align*} -3y^2 + 2y + 5 & \geq 0\\ 3y^2 - 2y - 5 & \leq 0\\ 3y^2 + 3y - 5y - 5 & \leq 0\\ 3y(y + 1) - 5(y + 1) & \leq 0\\ (3y - 5)(y + 1) & \leq 0 \end{align*} which holds if $y \in [-1,5/3]$. Therefore, the range of $$y = \frac{x^2 + 3x + 1}{x^2 + x + 1}$$ is $$\text{Ran} = \left[-1, \frac{5}{3}\right]$$

Answer 2

Points you should check are where $\dfrac{dy}{dx} = 0$ and limits of the range $\lim_{\pm\infty} \dfrac{dy}{dx}$

Answer 3

$$y=1+\frac{2x}{x^2+x+1}$$

The function $y(x)$ is continuous and differentiable in $\Bbb R$. Moreover $\lim_{x\to\pm\infty}y(x)=1$, so $y(x)$ is bounded. This means that the range of $y$ is an interval $I$.

Then $$y'=\frac{2x^2+2x+2-4x^2-2x}{(x^2+x+1)^2}=\frac{2-2x^2}{(x^2+x+1)^2}$$ which vanishes at $x=\pm1$.

Compute $y(1)$ and $y(-1)$ to find the endpoints of $I$.

Answer 4

$y = 1 +\frac{2x}{x^2 + x + 1} = 1 + \frac{2}{1+x+1/x}$, $x\ne 0.$

A) Let $x\gt 0:$

1) $x+1/x \ge 2$

AM-GM: $x+1/x \ge 2\sqrt{x×1/x}$

$y_{max} = 1 + \frac{2}{1+2} = 5/3$.

B) Let $x \lt 0:$

2) $x+1/x \le -2$

$y_{min} = 1 + \frac{2}{1- 2} = 1+ (-2) = -1.$

Since

$y = \frac{x^2 + 3x + 1}{x^2+x+1}$ is continuous on $\mathbb{R}$, we find:

Range$ = [-1,5/3].$