How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative?

Question

The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.

Maybe it can be solved by using inequality.

Answer 1

Does not exist.

Try $x\rightarrow-1^-$.

For $x>-1$ by AM-GM we obtain: $$\frac{x^2+1}{x+1}=\frac{x^2-1+2}{x+1}=x-1+\frac{2}{x+1}=$$ $$=x+1+\frac{2}{x+1}-2\geq2\sqrt{(x+1)\cdot\frac{2}{x+1}}-2=2\sqrt2-2.$$ The equality occurs for $x+1=\frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.

Thus, the range for $x>0$ it's $[2\sqrt2-2,+\infty).$

For $x<-1$ we can get the range by the similar way.

Answer 2

Since $$\lim_{x\to-\infty} \frac{x^2+1}{x+1} = -\infty,$$

the minimum does not exist.

Answer 3

The equation $$y={x^2+1\over x+1}=a\to x^2-ax-(a-1)=0$$for $x\ne -1$ has a real root if and only if $$\Delta\ge0 \iff a^2+4a-4\ge 0\iff a\ge 2\sqrt 2-2\text{ or }a\le -2\sqrt 2-2$$in that case, the root is $\ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-\infty,-2\sqrt 2-2]\ \bigcup \ [2\sqrt 2-2,\infty)$$

Answer 4

Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)

$y=\frac{x^2+1}{x+1}$ $$\Rightarrow yx + y = x^2 +1 \Rightarrow -x^2 + yx +(y-1)=0 \Rightarrow x = \frac{-y \pm\sqrt{y^2 + 4(y-1)}}{-2} \Rightarrow \frac{-y \pm \sqrt{(y-(-2-\sqrt5))(y - (\sqrt5-2))}}{-2}$$

The Lowest Value so will be obtained will be when the value under square root = 0, then in that case

$$ = \frac{-\sqrt5+2}{-2}$$

Now plug that value in the expression for $x$ Gives the minimal value of $y$: