Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

3psilon

  • 2 years ago

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

  • This Question is Closed
  1. 3psilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. 3psilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I never understood questions like this

  3. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -1/2?

  8. LolWolf
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yep.

  9. 3psilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So both limits approach from the left?

  10. LolWolf
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. 3psilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. LolWolf
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  15. LolWolf
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    (And, sure thing, of course)

  16. 3psilon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. LolWolf
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    All right, sounds good

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.