## 3psilon Group Title How to find a value continuous for all real numbers in this function? one year ago one year ago

1. 3psilon Group Title

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

2. 3psilon Group Title

I never understood questions like this

3. satellite73 Group Title

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

4. satellite73 Group Title

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

5. satellite73 Group Title

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 $$a$$

6. LolWolf Group Title

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.

7. 3psilon Group Title

-1/2?

8. LolWolf Group Title

Yep.

9. 3psilon Group Title

So both limits approach from the left?

10. LolWolf Group Title

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: $f(1)=x^2-a=(1)^2-a$

11. satellite73 Group Title

yeah i get $$-\frac{1}{2}$$

12. 3psilon Group Title

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

13. satellite73 Group Title

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

14. LolWolf Group Title

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

15. LolWolf Group Title

(And, sure thing, of course)

16. 3psilon Group Title

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

17. LolWolf Group Title

All right, sounds good