Question

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

Answer 1

\[ f(x) = \left\{ 2x + a, x < 1 \right\} \]
\[ f(x) = \left\{ x^{2} - a, x \ge 1 \right\} \]

Answer 2

I never understood questions like this

Answer 3

this is much much easier than you think
replace \(x\) by 1 on both expressions, set them equal, solve for \(a\)

Answer 4

that way they will agree at \(x=1\) and so it will be continuous there.
it is obviously continuous everywhere else

Answer 5

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 \(a\)

Answer 6

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.

Answer 7

-1/2?

Answer 8

Yep.

Answer 9

So both limits approach from the left?

Answer 10

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:
\[
f(1)=x^2-a=(1)^2-a
\]

Answer 11

yeah i get \(-\frac{1}{2}\)

Answer 12

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

Answer 13

@LolWolf gave a thorough and nice explanation. you have to admit it is easy right?

Answer 14

Haha, no, we were nailed on this stuff since our teacher is in love with definitions.

Answer 15

(And, sure thing, of course)

Answer 16

Yes crystal clear to me now. Gonna ace that calc test tomorrow now

Answer 17

All right, sounds good