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DHASHNI

  • 4 years ago

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

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  1. pratu043
    • 4 years ago
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    With what?

  2. DHASHNI
    • 4 years ago
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    |dw:1326694446525:dw|

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

  4. myininaya
    • 4 years ago
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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, \[\lim_{x \rightarrow \frac{\pi}{2}}f(x)=f(\frac{\pi}{2})\]

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

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

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

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

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

  10. myininaya
    • 4 years ago
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    or b

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

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

  13. DHASHNI
    • 4 years ago
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    k

  14. myininaya
    • 4 years ago
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    i think the question had a type-o

  15. imperialist
    • 4 years ago
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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, \[\lim_{x\to\pi/2}\frac{1-\sin^3x}{3\cos^2x}=\lim \frac{-3\sin^2x\cos x}{-6\cos x \sin x}= \lim_{x \to \pi/2} 0.5\sin x=0.5=a\]

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

  17. imperialist
    • 4 years ago
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    Similarly, solve for b.

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

  19. myininaya
    • 4 years ago
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    also*

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

  21. DHASHNI
    • 4 years ago
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    for b ?

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

  23. imperialist
    • 4 years ago
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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.

  24. myininaya
    • 4 years ago
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    i was trying to make the function continuous everywhere

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

  26. myininaya
    • 4 years ago
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    he knows

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

  28. imperialist
    • 4 years ago
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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.

  29. myininaya
    • 4 years ago
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    its already continuous at 0

  30. myininaya
    • 4 years ago
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    cos(0)=1

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

  32. myininaya
    • 4 years ago
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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

  33. DHASHNI
    • 4 years ago
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    k

  34. DHASHNI
    • 4 years ago
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    thanks guys !!!!!!!!!!!!!!!!

  35. myininaya
    • 4 years ago
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    np :)

  36. imperialist
    • 4 years ago
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    Glad to help :)

  37. DHASHNI
    • 4 years ago
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    ;)

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