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cpt

  • 2 years ago

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

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

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

  3. cpt
    • 2 years ago
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    What should I do after that? Add those two equations together?

  4. myininaya
    • 2 years ago
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    You will have a system of equations to solve just like from algebra class.

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

  6. myininaya
    • 2 years ago
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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?

  7. myininaya
    • 2 years ago
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    I mean equation err

  8. cpt
    • 2 years ago
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    oops you are right.

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

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

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

  12. myininaya
    • 2 years ago
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    It isn't approaching -2. It is approaching 2.

  13. myininaya
    • 2 years ago
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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.

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

  15. myininaya
    • 2 years ago
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    Have you ever solved a system of two linear equations?

  16. cpt
    • 2 years ago
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    Yes, like 4 years ago haha.

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

  18. myininaya
    • 2 years ago
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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.

  19. myininaya
    • 2 years ago
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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.

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

  21. myininaya
    • 2 years ago
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    Do you think you can rewrite equation 2 in that form?

  22. cpt
    • 2 years ago
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    32a - 4b = -8 ( 2nd equation right) right?

  23. myininaya
    • 2 years ago
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    looks good

  24. myininaya
    • 2 years ago
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    So we have 16a-4b=-16 32a-4b=-8

  25. myininaya
    • 2 years ago
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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.

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

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

  28. cpt
    • 2 years ago
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    I got the final answer of 6 for b. Is that right?

  29. cpt
    • 2 years ago
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    and .5 for a

  30. myininaya
    • 2 years ago
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    yep. good job.

  31. cpt
    • 2 years ago
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    yayyy, you should just be my tutor haha. Thank you so much :D

  32. myininaya
    • 2 years ago
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    np

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