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

1. 3psilon

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

2. 3psilon

I never understood questions like this

3. satellite73

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

4. satellite73

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

5. satellite73

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

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

-1/2?

8. LolWolf

Yep.

9. 3psilon

So both limits approach from the left?

10. LolWolf

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

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

12. 3psilon

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

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

14. LolWolf

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

15. LolWolf

(And, sure thing, of course)

16. 3psilon

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

17. LolWolf

All right, sounds good