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Roots of a polynomial and differenciaton

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data
    I read that if f'(x) is zero once in [a b] then f(x) has maximum two real roots.
    Why maximum? Shouldn't it be exactly 2?
    Or it has something to do with the case of repeated roots?

    2. Relevant equations



    3. The attempt at a solution
    was thinking as in figure
     

    Attached Files:

  2. jcsd
  3. Mar 28, 2012 #2
    Well in your case it has exactly two real roots, but in another case you might only have one real root or none. Try it out and draw a graph and see what you get. Often there would be one real root and one imaginary root.
     
  4. Mar 28, 2012 #3
    oh...like a quadritic equation with D=o , will have f'(x)=0 at -b/2a but it does not have any real roots ?
     
  5. Mar 28, 2012 #4
    Thanks :)
     
  6. Mar 28, 2012 #5
    Ok take for example x^2+x+1. Does this have any real roots? If yes then what are they?
     
  7. Mar 28, 2012 #6
    no it doesnt have any real roots. but f'(x)=0 at x=-0.5
    right?
     
  8. Mar 28, 2012 #7
    sry typing mistake i there meant D<0
     
  9. Mar 28, 2012 #8
    Yes you are correct. You take it's derivative, and you have to see where that derivative equals 0. Like in the equation i gave you 2x+1=0 implies that 2x+1=0 if and only if x=-0.5.
     
  10. Mar 28, 2012 #9

    Yes if you have the a negative discriminant, then you have two complex roots which are complex conjugates of one another.
     
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