Choose constants a and b so that the function is continuous everywhere!
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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Updated on April 06, 2020Comments
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Admin over 3 years
$$f(x) =\begin{cases} a-x, & \text{if $x \le -1$ } \\ \frac{x^2+2x+b}{x+1} & \text{if $x \gt -1$ } \\ \end{cases}$$
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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LazyGamer over 10 yearsThank you Marvis. How can I be sure that it is continuos everywhere? Shall I graf and see?
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LazyGamer over 10 years@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 $x=-1$.
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user3433489 over 10 yearsNo. 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: wolframalpha.com/input/…
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Dominic Michaelis over 10 yearsoh right sry $a=-1$ is right made a mistake
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user3433489 over 10 yearsnp. Thanks for helping Dominic! Have a great day.
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Dominic Michaelis over 10 years@Alex90 but the graph is not continuous, cause continuousis for functions not for graphs :)
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user3433489 over 10 yearsThe graph is not cont. because I have set a and b to 1. In this one I have set a=-1 and b=1. wolframalpha.com/input/…
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Dominic Michaelis over 10 years@Alex90 no you missunterstood, a graph can't be continuous at all, only a fucntion can be continuous
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user3433489 over 10 yearsAhhaaaa! :) Yes you are right!