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cpt Group Title

If f(x) is continuous and differentiable and f(x) = ... then b? f(x) = {ax^4 + 5x, x< or = 2 {bx^2 - 3x, x> 2, then b = ? Help pls

  • 11 months ago
  • 11 months ago

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  1. myininaya Group Title
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    1st set these two expressions equal to each other: \[\lim_{x \rightarrow 2^-}f(x)=\lim_{x \rightarrow 2^+}f(x)\]

    • 11 months ago
  2. myininaya Group Title
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    2nd set these two expressions equal to each other: \[\lim_{x \rightarrow 2^-}f'(x)=\lim_{x \rightarrow 2^+}f'(x)\]

    • 11 months ago
  3. cpt Group Title
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    What should I do after that? Add those two equations together?

    • 11 months ago
  4. myininaya Group Title
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    You will have a system of equations to solve just like from algebra class.

    • 11 months ago
  5. cpt Group Title
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    Well I got, 4a - 10 = 4b - 6 and then -32 + 5 = 4b - 3. Is that correct?

    • 11 months ago
  6. myininaya Group Title
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    The first equation on the left hand side you have 4a-10 I think it should be 16a+10. What happen to a in the second expression?

    • 11 months ago
  7. myininaya Group Title
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    I mean equation err

    • 11 months ago
  8. cpt Group Title
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    oops you are right.

    • 11 months ago
  9. myininaya Group Title
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    \[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\]

    • 11 months ago
  10. myininaya Group Title
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    \[4a(2)^3+5=2b(2)-3 \text{ second equation }\]

    • 11 months ago
  11. cpt Group Title
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    Wait isn't it a(-2)^4 + 5 (-2) since the limit is coming from the left side?

    • 11 months ago
  12. myininaya Group Title
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    It isn't approaching -2. It is approaching 2.

    • 11 months ago
  13. myininaya Group Title
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    You have the left is just approaching 2 from the left. You have the right is just approaching 2 from the right. Either way they are both still approaching 2 and not some other number.

    • 11 months ago
  14. cpt Group Title
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    Ahh ok I got it :) After I get these 2 equations then what should I do afterward?

    • 11 months ago
  15. myininaya Group Title
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    Have you ever solved a system of two linear equations?

    • 11 months ago
  16. cpt Group Title
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    Yes, like 4 years ago haha.

    • 11 months ago
  17. cpt Group Title
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    16a + 10 4b - 6 32a + 5 4b - 3 ( Can't I just substitute it and then solve it) ?

    • 11 months ago
  18. myininaya Group Title
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    You can put them both in ax+by=c form if you want. Like if we had something like 2x+3y=1 -2x+3y=1 This system is already set up for elimination. Add the equations together and solve for y first would be my strategy. 6y=2 y=2/6=1/3 Then plug into one of the other equations and solve for x.

    • 11 months ago
  19. myininaya Group Title
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    Or this system: 3x-6y=2 x+y=2 This one isn't set up for elimination. But you could multiply the bottom equation by -3 so we would have it setup for elmination. 3x-6y=2 -3x-3y=-6 ----------------add -9y=-4 y=4/9 and then so on to find x.

    • 11 months ago
  20. myininaya Group Title
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    \[a(2)^4+5(2)=b(2)^2-3(2) \text{ first equation }\] \[16a+10=4b-6\] \[16a-4b=-16 \text{ rewrote the equation 1} \]

    • 11 months ago
  21. myininaya Group Title
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    Do you think you can rewrite equation 2 in that form?

    • 11 months ago
  22. cpt Group Title
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    32a - 4b = -8 ( 2nd equation right) right?

    • 11 months ago
  23. myininaya Group Title
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    looks good

    • 11 months ago
  24. myininaya Group Title
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    So we have 16a-4b=-16 32a-4b=-8

    • 11 months ago
  25. myininaya Group Title
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    If we multiply the second or even the first (just not both of them) by -1, then we can set this up for elimination method.

    • 11 months ago
  26. myininaya Group Title
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    I'm trying to eliminate b. (you could go for a but I prefer working with smaller numbers)

    • 11 months ago
  27. myininaya Group Title
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    So -16a+4b=16 32a-4b=-8 ----------------add the equations together and what do we get?

    • 11 months ago
  28. cpt Group Title
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    I got the final answer of 6 for b. Is that right?

    • 11 months ago
  29. cpt Group Title
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    and .5 for a

    • 11 months ago
  30. myininaya Group Title
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    yep. good job.

    • 11 months ago
  31. cpt Group Title
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    yayyy, you should just be my tutor haha. Thank you so much :D

    • 11 months ago
  32. myininaya Group Title
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    np

    • 11 months ago
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