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Snowflake123

  • 2 years ago

Not sure how to do this: Please explain??? :D Write down the set of values of the constant k for which the equation 2X^3-3X^2-12X-7=k has exactly one real solution. Thanksss

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  1. SNSDYoona
    • 2 years ago
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    nice drawing :P

  2. Snowflake123
    • 2 years ago
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    isnt it finding a value for k????

  3. mukushla
    • 2 years ago
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    lol...tnx so for any real number \(k\) there is exactly one solution for the equation see the intersects of horizental lines and \(f(x)\)

  4. SNSDYoona
    • 2 years ago
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    why do u differentiate the equation ?

  5. mukushla
    • 2 years ago
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    |dw:1345805312784:dw|

  6. mukushla
    • 2 years ago
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    because i want to see what is the behaviour of function

  7. mukushla
    • 2 years ago
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    if it was something like this|dw:1345805467267:dw| for some values between \(k_1\) and \(k_2\) there is more than one real root

  8. mukushla
    • 2 years ago
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    so i differentiate it to see if there is some critical points like last drawing or not.

  9. Snowflake123
    • 2 years ago
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    so is the answer k=all real numbers?

  10. mukushla
    • 2 years ago
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    i think yes...if im not wrong

  11. mukushla
    • 2 years ago
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    oops : misread : \[f(x)=2x^3−3x^2−12x−7\]\[f'(x)=6x^2-6x-12=0\]it gives \(x^2-x-2=0\) so \(x=-1,2\) so we have two critical points

  12. mukushla
    • 2 years ago
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    sorry @Snowflake123 ...

  13. Mimi_x3
    • 2 years ago
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    you sub the x back into the y for k i think..

  14. mukushla
    • 2 years ago
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    exactly

  15. Snowflake123
    • 2 years ago
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    so whats the k values

  16. Mimi_x3
    • 2 years ago
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    why not you try and sub it back in

  17. Mimi_x3
    • 2 years ago
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    sub it into y = 2x^2 - 3x^2 - 12x - 7

  18. Mimi_x3
    • 2 years ago
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    2x^3**

  19. mukushla
    • 2 years ago
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    so the graph will be like this|dw:1345806482815:dw| just try to work it out

  20. Mimi_x3
    • 2 years ago
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    why are there two k's?

  21. mukushla
    • 2 years ago
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    the equation \(f'(x)\) has 2 roots so there are 2 k's we must work it out

  22. Mimi_x3
    • 2 years ago
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    according to my notes dont you find the roots from the stationary points? so there is only one line \(y=k\)

  23. mukushla
    • 2 years ago
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    for \(k=k_1,k_2\) there are exactly 2 real roots for the equation \(f(x)=k\) for \(k_1<k<k_2\) there are exactly 3 real roots for the equation \(f(x)=k\)

  24. mukushla
    • 2 years ago
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    so the region we lookin for is \[k>k_2 \ \text{&} \ k<k_1 \]

  25. mukushla
    • 2 years ago
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    what do u mean by '' there is only one line y=k''

  26. Mimi_x3
    • 2 years ago
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    woops forget it i made a mistake; youre right

  27. mukushla
    • 2 years ago
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    lol...i hope im right...

  28. Mimi_x3
    • 2 years ago
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    yes you are!! i just drew abit it differently so got kinda confused

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