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$$f(x) =\begin{cases} a-x, & \text{if $x \le -1$ } \\ \frac{x^2+2x+b}{x+1} & \text{if $x \gt -1$ } \\ \end{cases}$$

I try:
$a-x$ is continous everywhere for every $a$ when $x\le1$.
The other one however is continous only when $b=1$. Am I correct?

So now we should choose $a$ so that $f$ is continious everywhere. Which means:

$$\lim_{x\to -1^+}f(x)=f(-1)=\lim_{x\to -1^-}f(x)$$ As we approach from left: $$\lim_{x\to -1^-}f(x)=\lim_{x\to -1^-}a-x=a+1$$ $$f(-1)=a+1$$ As we approach from right: $$\lim_{x\to -1^+}f(x)=\lim_{x\to -1^+}\frac{x^2+2x+1}{x+1}=notDefined$$

Im stuck now! Can you please help? How may I continue? Is this the right way of solving this type of question? One thing that comes to my head is that $a+1=0$ so $a=-1$

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up vote 0 down vote accepted

You are almost right. Note that $$\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \dfrac{x^2+2x+1}{x+1} = \lim_{x \to -1^+} \dfrac{(x+1)^2}{x+1} = \lim_{x \to -1^+} x+1 = 0$$ (In fact this is why we choose $b=1$.)

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Thank you Marvis. How can I be sure that it is continuos everywhere? Shall I graf and see? – user31113 Apr 5 '13 at 19:12
@Alex90 A linear function (in your case $a-x$) is continuous everywhere. And so is a rational expression, except at zeros of the denominator. Hence, the only place where there could be a possible discontinuity is at $x=-1$. – user17762 Apr 5 '13 at 19:15

The function needs to be continuous, for $b=1$ the limit will surely exist, on the right hand side, now we only need to chose $a$ such that the limits are the same.

\begin{align*} \frac{x^2 +2x+b}{x+1}&=\frac{x^2+2x+1+b-1}{x+1}\\ &=\frac{x^2+2x+1}{x+1} +\frac{b-1}{x+1} \\ &=\frac{(x+1)^2}{x+1}+\frac{b-1}{x+1}\\ &=x+1+\frac{b-1}{x+1} \end{align*}

So we see the limit for $x\to -1$ only can exist when $b=1$. When $b=1$ we have $$\lim_{x\to -1} x+1=0$$ Hence $b=1$, and $a=-1$.

For $x\neq -1$ your funciton is polynomial and hence continuous.

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No. I just solved it as Marvis said. And I get a=-1 and b=1. And the graph is continous everywhere! a and b are not equal. Graph:… – user31113 Apr 5 '13 at 19:15
oh right sry $a=-1$ is right made a mistake – Dominic Michaelis Apr 5 '13 at 19:16
np. Thanks for helping Dominic! Have a great day. – user31113 Apr 5 '13 at 19:17
@Alex90 but the graph is not continuous, cause continuousis for functions not for graphs :) – Dominic Michaelis Apr 5 '13 at 19:17
The graph is not cont. because I have set a and b to 1. In this one I have set a=-1 and b=1.… – user31113 Apr 5 '13 at 19:30

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