# How to find the range of the function $\frac{x+2}{x+1}$ with domain $x \geq 0$?

How to find the range of the function $\frac{x+2}{x+1}$ with domain $x \geq 0$?

I am a high school student and stuck at this simple question on domains and ranges of functions. I have done the first part (when $x = 0$; $f(x) = 2$ therefore $f(x) \leq 2$), but can't continue. I know that the answer is $f(x) \gt 1$ because my book gives the following explanation: "For large values of $x$, $x+1 ≅ x+2$ therefore $f(x) \gt 1$"; is there any other simpler method (preferably algebraic) that will allow me to understand this answer better?

EDIT: corrected typing mistake

• No, it is $f(x)>1$. $f(x)$ can never equal $1$. $f(x)=1+\frac{1}{x+1}$ should make it clear $f(x)>1$. – user26486 Jun 27 '15 at 17:25
• Yes, you're right I changed it now – daljit97 Jun 27 '15 at 17:29
• Note that parantheses are very important - you wrote $x+2/x+1$ while you meant $(x+2)/(x+1)$. One answerer misinterpreted the question. – user26486 Jun 27 '15 at 17:35
• You are right, I should have used brackets to clarify better, however if parentheses are not used (following the basic rules of arithmetic) aren't we suppose to assume that those two expression are the same? – daljit97 Jun 27 '15 at 18:01
• $x+2/x+1=x+\frac{2}{x}+1\neq \frac{x+2}{x+1}$ – user26486 Jun 27 '15 at 18:03

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

Then rearranging terms to get $x$ in the form of $y$. $$yx+y=x+2$$ $$yx-x=2-y$$ $$x(y-1)=2-y$$ $$x=\frac{2-y}{y-1}$$ As $x \geq 0$ $$\frac{2-y}{y-1} \geq 0$$ $$\frac{y-2}{y-1} \leq 0$$ $$y \in (1,2]$$

• Thank you, I definitely understand it better with this approach. – daljit97 Jun 27 '15 at 17:35

Actually $1\color{red}{\lt} f(x)\le 2$ for $x\ge 0$.

To see this, note that we have $$\frac{x+2}{x+1}=\frac{(x+1)+1}{x+1}=1+\frac{1}{x+1}.$$ Here, note that this is strictly decreasing and that $\lim_{x\to\infty}\frac{1}{x+1}=0$.

take $x = 0$ then you have $$\frac{0+2}{0+1} = 2$$

Now take $x=1$ then you have $$\frac{3}{2} = 1.5$$

now as $x$ increases then the limit $$\lim_{x \to \infty} \frac{x+2}{x+1} = 1$$

You can see this by applying l'hopitals rule because we have the form of $\large{\frac{\infty}{\infty}}$ and so we are allowed to Differentiate up and down, to get $\frac{1}{1}$

However, as we are taking the limit. Then it will never actually be $1$ because we can never reach infinity.

so $y \in (1,2]$

• It is enough to observe $\frac{x+2}{x+1}=1+\frac{1}{x+1}$ and L'Hopital is overkill. – user26486 Jun 27 '15 at 17:33