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Show that Max|f(x)|>=1/4

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f(x) = x^3+ ax^2+ bx + c with a, b, c real. Show that

    M = max|f(x)|≥1/4
    -1 ≤ x ≤ 1
    and find all cases where equality occurs.

    2. Relevant equations


    3. The attempt at a solution

    |(-1 ≤ x^3 ≤ 1)+(-a ≤ ax^2 ≤ a)+(-b ≤ bx ≤ b)+c|≥1/4?
    not sure how to prove without knowing what a,b,c are?
     
  2. jcsd
  3. Feb 8, 2012 #2

    Mark44

    Staff: Mentor

    Adding inequalities (which are either true or false) doesn't make any sense.

    Basically, what you need to show is that the largest value of |f(x)| on the interval -1 <= x <= 1, is at least 1/4.

    Wouldn't the derivative be useful here?
     
  4. Feb 9, 2012 #3
    thanks
    so i have
    f '(x)=3x^2 + 2ax + b
    f ' (x) =0 to find critical values
    = -1/3a ± (√(4a^2 -12b))/6

    not entirely sure where to go from here
     
  5. Feb 9, 2012 #4

    Mark44

    Staff: Mentor

    1. When you start a line with =, you are saying that what is just above equals what's to the right of the = sign. In this case, you are saying that
    0 = -1/3a ± (√(4a^2 -12b))/6
    This is not true.
    When you solved using the quadratic formula, what did you solve for?
    2. There is an error what you did with the quadratic formula.

     
  6. Feb 9, 2012 #5
    sorry i meant x= (-2a±√(4a^2-12b))/6,
    not sure where I have gone wrong with the formula though (-b±√(b^2-4ac))/2a, correct?
    although im not too sure how this helps me
     
  7. Feb 9, 2012 #6
    im not meant to do this:

    (-1/3a ± (√(4a^2 -12b))/6)^3+ a(-1/3a ± (√(4a^2 -12b))/6)^2 +b(-1/3a ± (√(4a^2 -12b))/6)+c

    surly?
     
  8. Feb 9, 2012 #7

    Mark44

    Staff: Mentor

    That's better.

    So what are the conditions for which f'(x) = 0? Will there be no such x-values, 1 such x-value, or 2 separate x-values?
     
  9. Feb 9, 2012 #8

    Mark44

    Staff: Mentor

    No
    surly and surely are different words. Unless I miss my guess, you meant "surely."
     
  10. Feb 9, 2012 #9
    yes i meant surely, sorry.
    there will be two x values, assuming a,b≠0 i think
     
  11. Feb 9, 2012 #10
    im unsure of how to work out what these x values are when we dont know what a and b are.
    Is the second derivative useful once we have found out what the x values are to find the max
    f''(x)=6x+2a
     
  12. Feb 9, 2012 #11

    Mark44

    Staff: Mentor

    No, not true. If a = b = 2, there are no values for which f'(x) = 0.

    Like I asked before, what are the conditions for which f'(x) = 0? For what values of a and be will there be no critical points, 1 critical point, or 2 separate critical points?

    The Quadratic Formula gives you more information than you are using.

    Also, all you are given is that a and b are real. You can't assume they aren't zero.
     
  13. Feb 9, 2012 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The issue is not *just* whether there are two critical values of x, but whether or not these two values lie between -1 and +1. On the interval [-1,1], you may have: (i) no stationary point in (-1,1)(so the max/min occur at an endpoint x = +- 1); (ii) one stationary point in (-,1,1) and the other outside it (so the min occurs in (-1,1) and the max at +- 1, or vice-versa); (iii) two stationary points in (-1,1) (so that the max may occur in (-1,1) or at an endpoint, and similarly for the min.) Somehow you need to straighten out these possibilities.

    RGV
     
  14. Feb 9, 2012 #13
    the conditions are that the function has a maximum at x=n if f'(n)=0 and f''(0) is negative

    it wont have any critical points when
    -2a + √(4a^2 -12b)=0 so when a,b=0
    it will have two critical point when
    -2a - √(4a^2 -12b)=0 → b=2/3a^2 and a=√(3b/2)?
     
  15. Feb 9, 2012 #14
    how am i meant to work out if the critical values lie in between -1 and 1, without know what a and b are. Sorry it all getting quite confusing
     
  16. Feb 9, 2012 #15

    Mark44

    Staff: Mentor

    Not quite, but you're getting warmer. There won't be any critical points when the discriminant < 0. When will there be one critical point? Two distinct critical points?

    And as Ray pointed out, it makes a difference whether these points are in (-1, 1) or not.
     
  17. Feb 9, 2012 #16
    if the discriminant is > 0, the polynomial has two real roots, if the discriminant is = 0, the polynomial has one real root, and if the discriminant is < 0, the polynomial has no real roots
    so therefore
    if 4a^2-12b>0 it has two real roots so 0<4a^2-12b<=1,
    if 4a^2-12b=0 it has one real root so 4a^2-12b=0
    if 4a^2-12b<0 it has no real roots so -1<=4a^2-12b<0
    is this what you mean?
     
  18. Feb 9, 2012 #17
    when you say it makes a difference whether or not the points are in (-1,1) is this because its stated that x has to be in these points in the qu?
    or is it something i have missed?
    also what am i meant to do now, am i meant to solve for a and b?
     
  19. Feb 9, 2012 #18

    Mark44

    Staff: Mentor

    Yes and no, but mostly no.

    What has two real roots? Be specific. Don't use the word "it".
    Why do you think that 4a2 - 12b <= 1? Where did that come from?
    if 4a^2-12b=0, then of course if 4a^2-12b=0.
    Again, what has one real root?
    Why do you think that 4a2 - 12b >= -1?
     
  20. Feb 9, 2012 #19
    when i say it i mean the derivative of f'(x) or should i be working out the discriminant of f(x), i.e. b^2 c^2-4ac^3-4b^3d-27a^2d^2+18abcd
    so in our case would be
    a^2b^2-4b^3-4a^3-27c^2+18abc,

    and i realise what wrong with the <=1 and >=1, i was thinking about x.
     
  21. Feb 9, 2012 #20
    so therefore it woudl mean the polynomial
     
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