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Does f'(0)=0 where x in [-1,1] and f(x)<=f(0)

  1. Dec 7, 2015 #1
    1. The problem statement, all variables and given/known data
    If the function f:ℝ→ℝ is differentiable and f(x)<=f(0) for all x ∈[-1,1], then f'(0)=0. True or False.

    2. Relevant equations


    3. The attempt at a solution
    I think the statement is right. Since f(x)<= f(0) for all x in [-1,1], this tells us f is an even function or a symmetric function, then apply Rolle's theorem, we can prove the statement.
     
  2. jcsd
  3. Dec 7, 2015 #2

    andrewkirk

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    The function is not necessarily even or symmetric. Consider f(x) that is ##x^3## for ##x\leq 0## and ##-x^2## for ##x>0##.
     
  4. Dec 7, 2015 #3
    what is the sign of ##\frac{f(x) - f(0)}{x} ## for ##x## positive and negative ?
     
  5. Dec 7, 2015 #4

    Ray Vickson

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    Your function is differentiable on [-1,1] and has a maximum over [-1,1] at the point x = 0. What does that tell you?
     
  6. Dec 7, 2015 #5
    let h(x)=(f(x)-f(0))/x, when x>0, h(x)<=0; when x<0, h(x)=>0; f(x) is bounded above and the supremum is at x = 0, so f'(0)=0
     
  7. Dec 7, 2015 #6
    At x =0, we still have f'(0)=0?
     
  8. Dec 7, 2015 #7

    andrewkirk

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    Yes, the conclusion still holds. In the OP you reached the correct conclusion based on an incorrect argument. Ray's and Geoffrey's posts indicate the direction that a valid argument would take.
     
  9. Dec 8, 2015 #8
    Can you clearly explain the link between ##h(x)## and ##f'(0) ## ?


    EDIT: By the way, my hint is valid, but honestly, Mr Vickson's hint is more elegant
     
    Last edited: Dec 8, 2015
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