How to find the range of this expression

Question

I have :

$$ \frac{x^2+2x+1}{x^2+2x+7} $$

Where x is real.

I need to find out its range.

I tried:

Let $ \frac{x^2+2x+1}{x^2+2x+7} = y $

Which gives me: $ 24y^2 - 28y + 3 =0 $

And I am confused what to do next.

Answer 1

$$ \frac{x^2+2x+1}{x^2+2x+7}\geq0,$$ where the equality occurs for $x=-1$,

which says that $0$ is a minimal value. $$ \frac{x^2+2x+1}{x^2+2x+7}=1-\frac{6}{x^2+2x+7}<1$$ and since $$\lim_{x\rightarrow\infty}\frac{x^2+2x+1}{x^2+2x+7}=1,$$ we see that $1$ is a supremum and since our expression is continuous, we got the answer: $$[0,1)$$

Answer 2

$$ \frac{x^2+2x+1}{x^2+2x+7} =1-\frac{6}{x^2+2x+7}$$ Note that $x^2+2x+7=(x+1)^2+6\ge6$

So we have

$$\frac{6}{x^2+2x+7}=\frac{6}{(x+1)^2+6}\le 1$$

And $$1>1-\frac{6}{(x+1)^2+6}\ge0$$

$$1>\frac{x^2+2x+1}{x^2+2x+7}\ge0$$

Answer 3

hint : I prefer to use $$u=x+1$$ so you have $$\quad{y=\frac{x^2+2x+1}{x^2+2x+7}=\\\frac{(x+1)^2-1+1}{(x+1)^2-1+7}=\frac{u^2}{u^2+6}}$$now it is easy to find range $$\quad{y=\frac{u^2}{u^2+6}\to \\u^2y+6y-u^2=0\\u^2=\frac{-6y}{y-1}\\\text{ now } \frac{-6y}{y-1}\geq 0\\0\leq y<1}$$

Answer 4

We have $$ \frac{x^2+2x+1}{x^2+2x+7}= \frac{x^2+2x+7-6}{x^2+2x+7} =1-\frac{6}{x^2+2x+7}=1-\frac{6}{(x+1)^2+6}<1. $$ Also $$ \frac{x^2+2x+1}{x^2+2x+7}=\frac{(x+1)^2}{(x+1)^2+6} \geq 0. $$ Thus $$ 0 \leq \frac{x^2+2x+1}{x^2+2x+7} <1 $$