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Conclusions on conditions

  1. Oct 20, 2015 #1
    It is known that f"x>0,f(x)=f(-x),then which is correct?

    f(0)<f(1);
    f(4)-f(3)<f(6)-f(5);
    f(-2)<(f(-3)+f(-1))/2

    Can I simply use y=x^2 to conclude that only the third is incorrect?

    Thx!
     
  2. jcsd
  3. Oct 20, 2015 #2

    RUber

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    Homework Helper

    If y = x^2, then I don't see why the first would be wrong...0^2<1^2. Nor would the second be wrong 7 < 11.
    You may simply use one example as a counter example to prove something wrong, but that is not justification for them to always hold true.

    Use what you know.
    f'' (x) > 0, so you have a concave up function (like x^2).
    f(x) = f(-x) means you have symmetry about the y axis (like x^2).
    In order for this to happen, you must have a function that is decreasing in negative x and increasing in positive x. That is assuming that the function is continuously differentiable over the reals. If the derivative were not defined at zero, you could make a function that was increasing in the negative x and decreasing in the positive x.

    The second one property can be shown using the fact that a positive second derivative implies an increasing first derivative.

    The third property is a fundamental property of concavity. (f(-3)+f(-1)/2) is the linear average of the line connecting the points. That is, it would be the value of y(2) if y were the line connecting f(-1) with f(-3).
     
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