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JJ1371

  • 3 years ago

For what value of x does y=x-x^2 and y=x^2 have the same gradient?

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  1. apple_pi
    • 3 years ago
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    differentiate each and equate the derivatives

  2. akash123
    • 3 years ago
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    x=1/4

  3. apple_pi
    • 3 years ago
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    woah @akash123 where'd that come from?

  4. akash123
    • 3 years ago
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    differentiate each and equate the derivatives

  5. akash123
    • 3 years ago
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    :)

  6. apple_pi
    • 3 years ago
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    You were scaring me there

  7. apple_pi
    • 3 years ago
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    @JJ1371 taking it in...

  8. akash123
    • 3 years ago
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    where? proving the collinearity..

  9. apple_pi
    • 3 years ago
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    no, just suddenly pulling an answer

  10. akash123
    • 3 years ago
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    oh...ok

  11. JJ1371
    • 3 years ago
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    sorry, the second y=x^3

  12. apple_pi
    • 3 years ago
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    Ok, follow the procedure

  13. JJ1371
    • 3 years ago
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    y=1+2x y=2x^2 1+2x = 2x^2 where to now?

  14. akash123
    • 3 years ago
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    now u have to solve this quadratic equation and find the values of x

  15. akash123
    • 3 years ago
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    1+2x = 3x^2

  16. JJ1371
    • 3 years ago
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    okay 2x^2-2x-1=0

  17. akash123
    • 3 years ago
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    derivative of x^3 = 3 x^2

  18. JJ1371
    • 3 years ago
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    how did you get that?

  19. akash123
    • 3 years ago
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    derivative of x^n = n x^(n-1)

  20. JJ1371
    • 3 years ago
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    oh right yeah

  21. JJ1371
    • 3 years ago
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    sorry

  22. JJ1371
    • 3 years ago
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    3x^2 - 2x -1 = 0 Quadratic equation?

  23. JJ1371
    • 3 years ago
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    wait! (x-3)(x+1) = 0

  24. JJ1371
    • 3 years ago
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    therefore x = 3 or x = -1?

  25. akash123
    • 3 years ago
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    no...3x^2 - 2x -1 = 0 =(x-3)(x+1) is wrong do the factorization again

  26. JJ1371
    • 3 years ago
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    damn

  27. JJ1371
    • 3 years ago
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    x = 1 or x=-(1/3)

  28. JJ1371
    • 3 years ago
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    yes?

  29. akash123
    • 3 years ago
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    yes...:)

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