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anonymous

  • 5 years ago

The function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

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  1. anonymous
    • 5 years ago
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    so f ' (x) = 0 and f ' ' (x) = 0 for the same x . so f '(x) = f '' (x) f ' = 4x^3 + 2bx + 8 = f '' (x) = 12x^2 + 2b

  2. anonymous
    • 5 years ago
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    but that still doesn't give me the value of b... do i set it equal to 0?

  3. anonymous
    • 5 years ago
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    you have to solve that equation

  4. myininaya
    • 5 years ago
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    we have two equations 4x^3+2bx+8=0 and 12x^2+2b=0 I solve the bottom one for b and replace the b in the first equation with -6x^2

  5. anonymous
    • 5 years ago
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    4x^3 + 2bx + 8= 12x^2 + 2b

  6. anonymous
    • 5 years ago
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    oh thats smarter myininya has it

  7. anonymous
    • 5 years ago
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    my way involves a bit of work

  8. myininaya
    • 5 years ago
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    if you do this correctly you should get x=1 so now we can find b since b=-6x^2 then b=-6(1)^2=-6

  9. myininaya
    • 5 years ago
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    any questions?

  10. anonymous
    • 5 years ago
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    got it thank you!! :)

  11. myininaya
    • 5 years ago
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    :)

  12. anonymous
    • 5 years ago
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    i'm going to keep posting questions so feel free to answer :) lol... i have my ap test on wednesday so i'm reviewing a lot

  13. myininaya
    • 5 years ago
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    good luck

  14. anonymous
    • 5 years ago
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    thanks

  15. myininaya
    • 5 years ago
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    and you could have done it cantorset's way but it is a bit more thinking down his path

  16. myininaya
    • 5 years ago
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    i will show you if you want. i will do it on paper and scan it and post it

  17. anonymous
    • 5 years ago
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    I'll try working on it in a bit, but I like the way you taught me better, seems a lot easier.

  18. myininaya
    • 5 years ago
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  19. anonymous
    • 5 years ago
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    wow.. looks like a lot of work :)

  20. myininaya
    • 5 years ago
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    lol

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