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Find K for which f is one-to-one

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x)= x^3 + kx^2 + 19x


    3. The attempt at a solution

    So what i did was take the derivative of f(x)
    3x^2 + 2kx + 19
    \sqrt{}


    (3x+ [tex]\sqrt{19}[/tex]) (x+ [tex]\sqrt{19}[/tex])

    so then i can assume that 3x[tex]\sqrt{19}[/tex] + x[tex]\sqrt{19}[/tex] = 2kx

    and then k i got to be 2[tex]\sqrt{19}[/tex]

    Im i right?? i feel like i'm completely lost..
     
  2. jcsd
  3. Jan 23, 2009 #2

    Dick

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    f(x) is one to one if it's derivative never changes sign. f'(x)=3x^2+2kx+19 is always positive for large enough x, so you want to find the values k where f'(x) is never negative. Think about finding the minimum of f'(x) as a function of k.
     
  4. Jan 23, 2009 #3
    k so find where f(x) is increasing

    so..

    3x + [tex]\sqrt{19}[/tex] > 0

    x + [tex]\sqrt{19}[/tex] > 0

    So x> (-[tex]\sqrt{19}[/tex]) / 3
    and x> -[tex]\sqrt{19}[/tex]

    Is that right??
     
  5. Jan 23, 2009 #4

    Mark44

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    What happened to k? It shouldn't just go away.
     
  6. Jan 23, 2009 #5

    Mark44

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    No.
    Go back to the derivative you found and determine a value of k that makes the derivative positive everywhere. Instead of factoring, as you have attempted (unsuccessfully) to do, you might try the quadratic formula.
     
  7. Jan 23, 2009 #6
    k i used the quadratic formula and got

    [ -2k +/- [tex]\sqrt{2k^2 - 228}[/tex] ] / 6 = X

    Having trouble solving for k
     
  8. Jan 23, 2009 #7

    Dick

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    No, no, no. Find the minimum of f'(x). Take the derivative of f'(x) (i.e. f''(x)) and set it equal to zero. That will give you the x value of the point where f'(x) is a minimum. What is the value of f'(x) there?
     
  9. Jan 23, 2009 #8

    Mark44

    Staff: Mentor

    You have f'(x) = 3x^2 + 2kx + 19, and you would like to find a value of k so that f'(x) > 0 for all x.

    By using the Quadratic formula, so solved the equation 3x^2 + 2kx + 19 = 0, for x.
    If you want 3x^2 + 2kx + 19 to not have any (real) solutions for x, you want to pick a number k so that the discriminant (the part in the radical in your work) is negative. If that happens, there won't be any real solutions to the equation 3x^2 + 2kx + 19 = 0.

    What does it take for the discriminant to be negative?
     
  10. Jan 23, 2009 #9
    Taking the 2nd derivative i get 6x + 2k = 0
    so x=-2k/6


    take this values into the 1st derivative and i get a long equation that ends up giving me after reducing

    -k^2 - 2k/3 + 19
     
  11. Jan 23, 2009 #10

    Dick

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    You have that the minimum of f'(x) is at x=-k/3. That's good. But when I put x=-k/3 into 3x^2 + 2kx + 19 I DON'T get -k^2 - 2k/3 + 19. Could you fix that up?
     
  12. Jan 23, 2009 #11
    Opps i forgot the squared root sign.

    -k^2 - (2k^2)/3 + 19

    Do i set this to 0?
     
  13. Jan 23, 2009 #12

    Dick

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    I don't agree with the first term either. It was supposed to be 3*(-k/3)^2. Sure, once you get it right, then you can set it equal to zero and use that info to figure out where it's nonnegative.
     
  14. Jan 23, 2009 #13
    arg... okay i for some reason cancel the 3 out..

    new equation gave me k^2= 19
     
  15. Jan 23, 2009 #14

    Dick

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    Your algebra isn't doing so well today. 3*(-k/3)^2=k^2/3. Plus. Not minus.
     
  16. Jan 23, 2009 #15
    yea i think i got it.. half way thru the equation i figures 3x root19 + x root19
    need a minus sign
    i ma start all over.. give me a sec
     
  17. Jan 23, 2009 #16

    Dick

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    Don't start all over! You just need a good expression for the minimum of f'(x) in terms of k. Just fix the sign problem.
     
  18. Jan 23, 2009 #17
    No no i save the information on like 20 pages.. wanted to do it on one neat page for my records later on.

    I finally got k= +/- root 57
     
  19. Jan 23, 2009 #18

    Dick

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    Right. Those are the values of k for which f'(x) has a zero minimum. The minimum in general is 19-k^2/3. For what values of k is that nonnegative? +/-sqrt(57) aren't the only possible values of k.
     
  20. Jan 23, 2009 #19
    Thank You!!!

    I finally understand it

    so if im correct

    the function x^3 + kx^2 + 19x

    is one to one at
    k <= - root 57 OR root 57 <= k
     
  21. Jan 23, 2009 #20

    Dick

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    Other way around. -sqrt(57)<=k<=sqrt(57), right? For example, k=0 works.
     
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