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DHASHNI Group Title

help!!!!!!!!!!!!!!

  • 2 years ago
  • 2 years ago

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  1. pratu043 Group Title
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    With what?

    • 2 years ago
  2. DHASHNI Group Title
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    |dw:1326694446525:dw|

    • 2 years ago
  3. DHASHNI Group Title
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    if f(x)is continuous at x=0...............find a,b

    • 2 years ago
  4. myininaya Group Title
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    Ok so this means we want the function to have the following properties: \[f(\frac{\pi}{2}) \text{ exist}\] \[\lim_{x \rightarrow \frac{\pi}{2}}f(x) \text{ exist}\] Finally, \[\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})\]

    • 2 years ago
  5. myininaya Group Title
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    so this means also we need \[\lim_{x \rightarrow \frac{\pi}{2}^+}f(x)=\lim_{x \rightarrow \frac{\pi}{2}^-}f(x)\]

    • 2 years ago
  6. myininaya Group Title
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    so for x=pi/2 f(x)=a we will use this later!

    • 2 years ago
  7. DHASHNI Group Title
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    myini: the question says that f(x) us continuous at x=0

    • 2 years ago
  8. myininaya Group Title
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    \[f(\frac{\pi}{2})=a\]

    • 2 years ago
  9. myininaya Group Title
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    oh that's interesting we wouldn't care about a then

    • 2 years ago
  10. myininaya Group Title
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    or b

    • 2 years ago
  11. myininaya Group Title
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    because that part says x>pi/2

    • 2 years ago
  12. myininaya Group Title
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    0 is less than pi/2

    • 2 years ago
  13. DHASHNI Group Title
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    k

    • 2 years ago
  14. myininaya Group Title
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    i think the question had a type-o

    • 2 years ago
  15. imperialist Group Title
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    Just take a couple of limits using l'hopitals rule. For the first one since \[1-\sin^3(\pi/2)=3\cos^2(\pi/2)=0\] the constraints of l'hopital's rule applies for the left limit. Thus, \[\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}=\lim \frac{-3\sin^2x\cos x}{-6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a\]

    • 2 years ago
  16. myininaya Group Title
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    because there is nothing to do if they aren't talking about pi/2

    • 2 years ago
  17. imperialist Group Title
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    Similarly, solve for b.

    • 2 years ago
  18. myininaya Group Title
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    and see imperialist ignored that continuous part at 0 lol

    • 2 years ago
  19. myininaya Group Title
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    also*

    • 2 years ago
  20. myininaya Group Title
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    but yeah i think they meant at pi/2

    • 2 years ago
  21. DHASHNI Group Title
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    for b ?

    • 2 years ago
  22. imperialist Group Title
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    Yeah, me too, I didn't even notice the zero part.

    • 2 years ago
  23. imperialist Group Title
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    For b, you just do the same thing, except you take the limit as x approaches pi/2 from the right regarding the second half of the piecewise function. Set that equal to 1/2, since you know the limit must be that, and you are done.

    • 2 years ago
  24. myininaya Group Title
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    i was trying to make the function continuous everywhere

    • 2 years ago
  25. DHASHNI Group Title
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    @imper:but the question says that the function is continuous at x=0

    • 2 years ago
  26. myininaya Group Title
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    he knows

    • 2 years ago
  27. imperialist Group Title
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    It's a typo, it's impossible to make it continuous at x=0.

    • 2 years ago
  28. imperialist Group Title
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    You already know how x is defined at zero, that can't be changed, no matter what you make a or b.

    • 2 years ago
  29. myininaya Group Title
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    its already continuous at 0

    • 2 years ago
  30. myininaya Group Title
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    cos(0)=1

    • 2 years ago
  31. imperialist Group Title
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    True dat, my bad, I didn't look back at the question and assumed sin was on the bottom.

    • 2 years ago
  32. myininaya Group Title
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    thats why the question makes no sense the work is already done i believe it meant what me and imperialist were trying to do

    • 2 years ago
  33. DHASHNI Group Title
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    k

    • 2 years ago
  34. DHASHNI Group Title
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    thanks guys !!!!!!!!!!!!!!!!

    • 2 years ago
  35. myininaya Group Title
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    np :)

    • 2 years ago
  36. imperialist Group Title
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    Glad to help :)

    • 2 years ago
  37. DHASHNI Group Title
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    ;)

    • 2 years ago
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