Finding the range of $\frac{x+1}{x^2+2}$

Question

There is a metheord to find a range of a some rational function $f$

For example think the function $\frac{x+1}{x^2+2}$

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

thus $y x^2-x+2y-1=0$

since this need to give real $x$,

$\Delta_x=b^2-4ac=1-4y(2y-1)\geq 0$

thus $\frac{1-\sqrt 3}{4} \leq y \leq \frac{1+\sqrt 3}{4}$

thus range is $[\frac{1-\sqrt 3}{4} , \frac{1+\sqrt 3}{4}]$

But I think from this we can't say that range is exactly the above result

Only we can get the fact that range is a sub-set of the above interval, if we need to prove range is equal to the interval we need to use the continuity and maximum and minimum values of $y$ when $x\in \Bbb R$

Am I right? What are your thoughts? Thanks.

Answer 1

Well it should be $\Delta _x\geq 0$, so $$-8y^2+4y+1\geq 0$$

and thus $y\in [y_1,y_2]$ where $y_i$ are solution of $-8y^2+4y+1= 0$, so your solution is correct.

Answer 2

Hint

Let $f(x)=\frac{x+1}{x^2+2}$. $$\lim_{x\to \pm \infty }f(x)=0.$$

Find the extrema of the function an you'll have your answer.