Is y=x^2 enough to prove the third statement incorrect?

[Zhang Jiawen]

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!

 

[RUber]

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).