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Finding the number of rational values a function can take

  1. Aug 5, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    ##f(x)## is a continuous and differentiable function. ##f(x)## takes values of the form ##^+_-\sqrt{I}## whenever x=a or b, (where ##I## denotes whole numbers) ; otherwise ##f(x)## takes real values. Also, ##|f(a)|\le |f(b)|## and ##f(c)=-1.5##. Graph of ##y=f(x)f'(x)##:
    Untitled.png
    The number of rational values that ##f(a)+f(b)+f(c)## can take is?

    2. Relevant equations
    None

    3. The attempt at a solution
    My teacher told me that the question has data insufficiency. But I dont know why.
    ##f(x)f'(x)=0## implies either ##f(x)## or ##f'(x)## or both are zero. At ##c## ##f'(x)## is zero since ##f(x)## is non zero. Hence, f(x) has a maxima or minima at ##x=c##.
    Now, from ##a## to ##c##, either bothe the function and its derivative is positive or both are negative. Applying Rolle's theorem,
    ##{f'(k)}^2+f(k)f"(k)=0## since ##g(a)=(c)##.
    But I dont seem to be getting anywhere.
     
  2. jcsd
  3. Aug 5, 2015 #2

    RUber

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    Looking at the plot, between a and c, f(x) and f'(x) are either both negative or both positive. Since f(c) = -1.5, we should assume both negative. Then from b to c, f(x) and f'(x) should have different signs, and we know that f(c) is negative, so assume f(x) <0 f'(x) >0.
    So the range of f(a) is bounded, as is the range of f(b). And you know that they are integers.
    f(c) is fixed, so it will not contribute to the number of possible solutions to f(a) + f(b) + f(c).
    So, how many whole numbers can f(a) be? what about f(b)?
     
  4. Aug 5, 2015 #3

    Titan97

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    f(a) and f(b) is always irrational. (given in the question). So, for f(a) + f(b) + f(c) to be rational, f(a)=-f(b). Also, f(a)<0 and f(b)>0.
     
  5. Aug 5, 2015 #4

    RUber

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    The question says that f(a) and f(b) are roots of a whole number. So 1 and 0 are not out of the question.
    if f(b) > 0, where does the function cross the axis, i.e. f(x)=0?
     
  6. Aug 5, 2015 #5

    SammyS

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    If ##\ I\ ## is a perfect square, then ##\ \sqrt{I}\ ## is rational.
     
  7. Aug 5, 2015 #6

    RUber

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    On interval (a,c), f(x) is either negative and decreasing or positive and increasing; which option makes it possible to continuously arrive at f(c) = -1.5?
    On interval (c,b), f(x) is either negative and increasing or positive and decreasing; which option makes it possible to begin continuously from f(c) = -1.5?
    Is it possible for there to be any value in (c,b) where f(x) = 0? What would the plot look like at that point?
    What whole numbers have roots in your possible ranges?
    What combinations of those values are possible after imposing ##|f(a)| \leq |f(b)|##?
    I get a number less than 10.
     
  8. Aug 6, 2015 #7

    Titan97

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    f(c)<0. This means, f'(c)<0. This is only for x=c. How can you say its true for all ##x\in [a,c]##?
     
  9. Aug 6, 2015 #8

    SammyS

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    The function is continuous, right.

    If it changes sign to the left of c, then what value must it pass through?
     
  10. Aug 6, 2015 #9

    RUber

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    You mean f'(c) = 0, as you said in your initial post.
    Between a and c, neither f nor f' can be zero, as SammyS mentioned, so your possible values for f(a) are constrained.
    Similarly, between c and b, neither f nor f' can be zero, so you are constrained for possible values at f(b).
    Add in the additional constraints of |f(a)| ≤ |f(b)| and then counting only the rational sums does not leave many possible options.
     
  11. Aug 7, 2015 #10

    Titan97

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    I have been trying this problem for hours.
    f'(c)=0, f(x)f'(x)>0 for ##x\in [a,c]##. But f(c)<0. So f(x) between a and c is also negative since function is continuous. But can't f(x) have a root between a and c?
     
  12. Aug 7, 2015 #11

    SammyS

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    What would that imply for ##\ y=f(x) f'(x) \ ## at that root ?
     
  13. Aug 7, 2015 #12

    Titan97

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    it implies that y=0
     
  14. Aug 7, 2015 #13

    SammyS

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    Does that shed some light on this problem?
     
  15. Aug 7, 2015 #14

    Titan97

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    I am confused. Here are some plots for f(x) which are possible:
    Untitled.png

    Untitled2.png

    Untitled3.png
     
  16. Aug 7, 2015 #15

    SammyS

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  17. Aug 7, 2015 #16

    Titan97

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    Why cant there be a root? The graphs are for y=f(x) and not f(x)f'(x). There are no roots for y=f(x)f'(x) in a<x<c
     
  18. Aug 7, 2015 #17

    haruspex

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    None of those look right to me. The curve given in the OP implies that in (a,c) f and f' have the same sign and in (c,b) they have the opposite sign.
    But I do count three possibilities for the shape of f.
     
  19. Aug 8, 2015 #18

    Titan97

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    Ok. I will analyse the curve separately from a to c and from c to b.
    From a to c:
    From ##a## to ##c##, f(x) and f'(x) have same sign.
    Since ##y=f(x)f'(x)## is not zero for any ##x\in (a,c)##, ##f(x)## and ##f'(x)##≠0. At ##x=a##, ##f(a)f'(a)=0##. So either ##f(a)## or ##f'(a)## or both are zero. giving 3 possibilities.
    Since ##y>0## from ##a## to ##c##, and ##f(c)<0##, ##f'(x)## and ##f(x)## is forced to be negative. ##f(a)## may be zero.
    So the curve can be either red or green:
    Untitled.png
     
  20. Aug 8, 2015 #19

    haruspex

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    Yes, but you could make it clearer that the red curve isn't necessarily zero gradient at a. It need not have an inflexion between a and c.
     
  21. Aug 8, 2015 #20

    Titan97

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    ok. It could be zero. From c to b:
    1. y=f(x)f'(x)<0 so f(x) and f'(x) have opposite signs
    2. since f(c)<0, f'(x)>0 in (c,b) hence an increasing curve.
    3. Again, y≠0 in (c,b) so the curve is entirely below/above x-axis.
    4. f'(c)=0 and f(b)f'(b)=0.
    So here are the possible curves:
    Untitled.png
     
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