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3psilon Group Title

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

  • one year ago
  • one year ago

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  1. 3psilon Group Title
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    \[ f(x) = \left\{ 2x + a, x < 1 \right\} \] \[ f(x) = \left\{ x^{2} - a, x \ge 1 \right\} \]

    • one year ago
  2. 3psilon Group Title
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    I never understood questions like this

    • one year ago
  3. satellite73 Group Title
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    this is much much easier than you think replace \(x\) by 1 on both expressions, set them equal, solve for \(a\)

    • one year ago
  4. satellite73 Group Title
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    that way they will agree at \(x=1\) and so it will be continuous there. it is obviously continuous everywhere else

    • one year ago
  5. satellite73 Group Title
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    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 \(a\)

    • one year ago
  6. LolWolf Group Title
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    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.

    • one year ago
  7. 3psilon Group Title
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    -1/2?

    • one year ago
  8. LolWolf Group Title
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    Yep.

    • one year ago
  9. 3psilon Group Title
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    So both limits approach from the left?

    • one year ago
  10. LolWolf Group Title
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    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: \[ f(1)=x^2-a=(1)^2-a \]

    • one year ago
  11. satellite73 Group Title
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    yeah i get \(-\frac{1}{2}\)

    • one year ago
  12. 3psilon Group Title
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    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

    • one year ago
  13. satellite73 Group Title
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    @LolWolf gave a thorough and nice explanation. you have to admit it is easy right?

    • one year ago
  14. LolWolf Group Title
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    Haha, no, we were nailed on this stuff since our teacher is in love with definitions.

    • one year ago
  15. LolWolf Group Title
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    (And, sure thing, of course)

    • one year ago
  16. 3psilon Group Title
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    Yes crystal clear to me now. Gonna ace that calc test tomorrow now

    • one year ago
  17. LolWolf Group Title
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    All right, sounds good

    • one year ago
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