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Minimum number of real roots

  1. Oct 16, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let f(x) be a non-constant twice differentiable function defined on R such that f(x)=f(1-x) and f'(1/4) =0 then what is the minimum number of real roots of the equation (f"(x))^2+f'(x)f'''(x)=0.

    3. The attempt at a solution

    f'(x)=-f'(1-x)
    f"(x)=f"(1-x)
    f'''(x)=-f'''(1-x)

    putting x = 1/4 in the first eqn gives f'(3/4)=0
    putting x=1/4 in the given equation (f"(x))^2+f'(x)f'''(x)=0 gives f"(1/4)=0. ∴f"(3/4)=0.
    So I have got a total of 2 roots. But there are some more which I can't find.
     
  2. jcsd
  3. Oct 17, 2013 #2
    Observe that the given equation is
    $$\frac{d}{dx}(f'(x)\cdot f''(x))=0$$
    See if that helps.
     
  4. Oct 17, 2013 #3

    utkarshakash

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    Let y= f'(x)f"(x)

    According to Rolle's Theorem there must lie a root of dy/dx between 1/4 and 3/4. But what about other roots?
     
  5. Oct 17, 2013 #4
    I am unsure how to proceed but how about x=1/2?

    At how many points is f'(x) zero? At how many points is f''(x) zero?
     
    Last edited: Oct 17, 2013
  6. Oct 17, 2013 #5

    pasmith

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    You need to find the known zeroes of [itex]y = f'f''[/itex], at which point you can apply Rolle's theorem to get the known zeroes of [itex]y'[/itex], which is what you're interested in.

    So far you know that [itex]f'(\frac14) = f'(\frac34) = 0[/itex].

    What, in view of the condition [itex]f(x) = f(1-x)[/itex], can you say about [itex]f'(\frac12)[/itex]?

    Now apply Rolle's theorem to [itex]f'[/itex].
     
  7. Oct 18, 2013 #6

    utkarshakash

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    I can say that there must lie a root of y' between 1/4 and 1/2 as well as 1/2 and 3/4.
     
  8. Oct 18, 2013 #7
    y is also zero at some more points.

    f'(x) is zero at 1/2 and 1/4 so there must be at least one point in between them where f''(x) is zero. Can you proceed now?
     
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