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anonymous

  • one year ago

which points are the best approximations of the relative maximum and minimum of the function f(x)=x^3+3x^2-9x-8

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  1. anonymous
    • one year ago
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    theser are the answer options HELP!!

  2. amoodarya
    • one year ago
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    find f' find roots of f'=0 check them by f' sign table or ,sign of f'' to find out max or min

  3. anonymous
    • one year ago
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    find f'(x) and then equate to 0 to find the critical points

  4. anonymous
    • one year ago
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    then check f"(x) at each point.

  5. anonymous
    • one year ago
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    would it be B??

  6. anonymous
    • one year ago
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    i really need please pls someone

  7. anonymous
    • one year ago
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    we have not solved yet. you solve t and show ,we will guide you.

  8. anonymous
    • one year ago
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    i just dont even know where to start..

  9. anonymous
    • one year ago
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    differentiate with respect to x

  10. anonymous
    • one year ago
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    \[f \prime \left( x \right)=3x^2+6x+9\]equate it to zero and find values of x

  11. anonymous
    • one year ago
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    correction it is -9 not +9

  12. anonymous
    • one year ago
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    so f(x) = 9 when equated to 0??

  13. anonymous
    • one year ago
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    no \[f \prime \left( x \right)=0~means~3x^2+6x-9=0\] divide by 3

  14. anonymous
    • one year ago
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    oh so x^2 + 2x - 3 = 0

  15. anonymous
    • one year ago
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    make factors

  16. anonymous
    • one year ago
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    x^2+(3-1)x-3=0 x^2+3x-1x-3=0

  17. anonymous
    • one year ago
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    how do u make factors

  18. anonymous
    • one year ago
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    3-1=-3 3*-1=-3

  19. anonymous
    • one year ago
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    correction 3-1=2

  20. anonymous
    • one year ago
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    coefficient of x^2*constant term is same as the product of two terms (made from coefficient of x)

  21. anonymous
    • one year ago
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    ok i still dont understand ugh

  22. anonymous
    • one year ago
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    coefficient of x^2=1 constant term=-3 product=1*-=-3 if the product is negative always subtract something to make the middle term here 2=3-1 product of 3*-1=-3

  23. anonymous
    • one year ago
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    i am leaving for sometime . i will be back to help you after 5 minutes.

  24. anonymous
    • one year ago
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    ok

  25. anonymous
    • one year ago
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    nevermind i figured it out

  26. anonymous
    • one year ago
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    i am here.

  27. anonymous
    • one year ago
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    have you found f"(x)

  28. anonymous
    • one year ago
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    Have you calculated critical points? \[f \prime \prime \left( x \right)=6x+6\]

  29. anonymous
    • one year ago
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    x(x+3)-1(x+3)=0 (x+3)(x-1)=0 x+3=0,x=-3 or x-1=0,x=1

  30. anonymous
    • one year ago
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    \[f \prime \prime \left( -3 \right)=?\]

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