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3psilon
 4 years ago
How to find a value continuous for all real numbers in this function?
3psilon
 4 years ago
How to find a value continuous for all real numbers in this function?

This Question is Closed

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

3psilon
 4 years ago
Best ResponseYou've already chosen the best response.0I never understood questions like this

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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 \(a\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

3psilon
 4 years ago
Best ResponseYou've already chosen the best response.0So both limits approach from the left?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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: \[ f(1)=x^2a=(1)^2a \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i get \(\frac{1}{2}\)

3psilon
 4 years ago
Best ResponseYou've already chosen the best response.0OHHH 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, no, we were nailed on this stuff since our teacher is in love with definitions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(And, sure thing, of course)

3psilon
 4 years ago
Best ResponseYou've already chosen the best response.0Yes crystal clear to me now. Gonna ace that calc test tomorrow now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0All right, sounds good
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