Here's the question you clicked on:
3psilon
How to find a value continuous for all real numbers in this function?
\[ f(x) = \left\{ 2x + a, x < 1 \right\} \] \[ f(x) = \left\{ x^{2} - a, x \ge 1 \right\} \]
I never understood questions like this
this is much much easier than you think replace \(x\) by 1 on both expressions, set them equal, solve for \(a\)
that way they will agree at \(x=1\) and so it will be continuous there. it is obviously continuous everywhere else
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 \(1-a\) so write \[2+a=1-a\] and solve for \(a\)
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: \[ \lim_{x\to 1^-} f(x)=\lim_{x\to 1^-} 2x+a=f(x)=(1)^2-a \]Using this, we get: \[ 2(1)+a=1-a \]Solve for a.
So both limits approach from the left?
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 set-up should be, if the limit exists: \[ \lim_{x\to 1^-}2x+a=\lim_{x\to 1^+}x^2-a=f(1) \]Where: \[ f(1)=x^2-a=(1)^2-a \]
yeah i get \(-\frac{1}{2}\)
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
@LolWolf gave a thorough and nice explanation. you have to admit it is easy right?
Haha, no, we were nailed on this stuff since our teacher is in love with definitions.
(And, sure thing, of course)
Yes crystal clear to me now. Gonna ace that calc test tomorrow now