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Homework Help: Derivative Applications

  1. Jan 14, 2007 #1
    I am stuck on the following questions lately. I hope someone can explain/help me out, any little help is greatly appreciated! :smile:

    1) Let f(x) = x^3 - 3x + b.
    1a) Show that f(x)=0 for at most one number x in [-1,1].
    1b) Determine the values of b such that f(x)=0 for some x in [-1,1].

    This is an example in my textbook,
    For part a, they showed f'(x)=0 at x=+/-1, so f'(x) not=0 on (-1,1), so f(x) = 0 has at most one root (otherwise, by rolle's theorem,....)
    For part b: they said that when f(x)=0, b = 3x - x^3 = x (3 - x^2). When x E [-1,1], |x (3-x^2)|<=2. Thus |b|<=2
    "When x E [-1,1], |x (3-x^2)|<=2" <------This is where I can't follow this logic, can someone please explain this?

    2) Let f and g be differentiable functions on the interval (0,c) such that f(0)=g(0). Prove that if f'(x)>g'(x) for all x E (0,c), then f(x)>g(x) for all x E (0,c).

    My attempt:
    Let F(x)=f(x)-g(x), I need to prove that F(x)>0
    F'(x) = f'(x)-g'(x) > 0 on x E (0,c) since f'(x)>g'(x)
    So F increases on (0,c)
    This is not what I want...how can I actually prove that F(x)>0?

    3) Determine A and B so that the graph of f(x) = Acos(2x) + Bsin(3x) will have a point of inflection at (pi/6,5).

    My solution: (incompleted)
    (pi/6,5) must lie on the curve, so I sub. (pi/6,5) into f(x), getting A+2B=10 ...........(1)

    f''(x) exists for all x E R, if there is a point of inflection at x=pi/6, then it must be true that f''(pi/6)=0, so I set f''(pi/6)=0, obtaining -2A-9B=0 .........(2)

    Solve (1) and (2), I get A=18,B=-4

    BUT the most important thing about point of inflection is that f''(x) MUST change sign (change concavity),f''(x)=0 does not mean that a certain point is a point of inflection for sure, so now I must show that in a small neighbourhood around x=pi/6, f'' has opposite sign, sub. A=18,B=-4 into f''(x), I get f''(x)=-36[2cos(2x)-sin(3x)], but now how can I prove that the 2nd derivative will change sign at x=pi/6? The trouble is that I don't know the location of the other inflection points, so I can't really use the "test sign" method (because I don't know in what interval will the sign of f'' keep constant), say if I evalutate f''(0) to test sign, the trouble is that between 0 and pi/6, f'' may not be keeping a constant sign.......how can I take care of this mess?
  2. jcsd
  3. Jan 14, 2007 #2


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    I'll have a go at 1:

    If you look at x(3-x2), considering we are in the range [-1,1], what are the maximum and minimum values of this expression?
  4. Jan 14, 2007 #3


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    To show if a critical point is a point of inflection, look at the values of f'(x) either side of the critical point. If the are opposite in sign, then you have a point of inflection.
  5. Jan 14, 2007 #4


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    2) You also know the value of F(0).
  6. Jan 14, 2007 #5
    How can I calculate this? The maximum and minimum may not occur at the endpoints......
  7. Jan 14, 2007 #6
    Yes, I know this fact perfectly, but my problem with question 2 I don't think it's possible to do so without knowing ALL the points where the 2nd derivative may possibly change sign...
    Last edited: Jan 14, 2007
  8. Jan 14, 2007 #7
    (i) F'(x) = f'(x)-g'(x) > 0 on x E (0,c)
    (ii) F increases on the OPEN interval (0,c)
    (iii) F(0)=f(0)-g(0)=0

    Do the 3 conditions above guarantee that F(x)>0 for all x E (0,c)? What if F is discontinuous at 0? (such as jump discontinuity), my thought is that this won't mean that F(x)>0 for all x E (0,c), is there any missing information in the actual question itself?
  9. Jan 17, 2007 #8
    Can anyone explain a little bit further? Any help is appreciated!
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