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How to find a value continuous for all real numbers in this function?
 one year ago
 one year ago
How to find a value continuous for all real numbers in this function?
 one year ago
 one year ago

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3psilonBest ResponseYou've already chosen the best response.0
\[ f(x) = \left\{ 2x + a, x < 1 \right\} \] \[ f(x) = \left\{ x^{2}  a, x \ge 1 \right\} \]
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
I never understood questions like this
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
this is much much easier than you think replace \(x\) by 1 on both expressions, set them equal, solve for \(a\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
that way they will agree at \(x=1\) and so it will be continuous there. it is obviously continuous everywhere else
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you can almost do it in your head if you replace \(x\) by 1 in the first expression you get \(2+a\) and in the second you get \(1a\) so write \[2+a=1a\] and solve for \(a\)
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
Well, we know that a function is continuous if at a point iff (they both must exist): \[ \lim_{x\to a} f(x)=f(x) \]So, we know both functions are continuous given their bounds, so if we use the limit definition: \[ \lim_{x\to 1^} f(x)=\lim_{x\to 1^} 2x+a=f(x)=(1)^2a \]Using this, we get: \[ 2(1)+a=1a \]Solve for a.
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
So both limits approach from the left?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
Well, we know the limit approaching from the right *is* \(f(x)\), so I didn't add that since I was trying to answer the question given some time, but the setup should be, if the limit exists: \[ \lim_{x\to 1^}2x+a=\lim_{x\to 1^+}x^2a=f(1) \]Where: \[ f(1)=x^2a=(1)^2a \]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
yeah i get \(\frac{1}{2}\)
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
OHHH I see why they are equal now! Its so clear ! Thanks guys! @LolWolf do you read the calculus book in your spare time or something ? Lol thanks
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
@LolWolf gave a thorough and nice explanation. you have to admit it is easy right?
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
Haha, no, we were nailed on this stuff since our teacher is in love with definitions.
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
(And, sure thing, of course)
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
Yes crystal clear to me now. Gonna ace that calc test tomorrow now
 one year ago

LolWolfBest ResponseYou've already chosen the best response.2
All right, sounds good
 one year ago
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