Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

3psilon

  • 3 years ago

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

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

    I never understood questions like this

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

    -1/2?

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

    Yep.

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

    So both limits approach from the left?

  10. LolWolf
    • 3 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. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

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

    (And, sure thing, of course)

  16. 3psilon
    • 3 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
    • 3 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

Sign Up
Find more explanations on OpenStudy
Privacy Policy