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Let a function f(x) be twice differentiable
such that $ f(0) = 0, f(\pi/2)= 1 , f(3\pi/2)=-1$. To prove that there exists a ‘c’ in $ (0,3{\pi/2})$ such that |$ f”(x) $ | is less than or equal to 1.

my conjecture is that question is wrong. by given info i cant prove the above condition. I have just one info about one point where f’ will be zero by rolle theorem. to comment on f” i need one more root of f’. need to confirm whether i am right.

maveric
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  • Perhaps $g(x)=f(x)-\sin(x)$ then $g(0)=g(\pi/2)=g(3\pi/2)=0$. Not sure if this is helpful yet. – kingW3 Sep 12 '19 at 19:37

1 Answers1

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If we take $g(x)=f(x)-\sin(x)$ then $g(0)=g(\pi/2)=g(3\pi/2)$ so we have that there exists $c_1 \in (0,\pi/2),c_2\in(\pi/2,3\pi/2)$ such that $$g'(c_1)=0=g'(c_2)$$ so there exists $c_3\in(c_1,c_2)$ such that $g''(c_3)=0$ but $$g''(c_3)=f''(c_3)+\sin(c_3)=0$$ so $$|f''(c_3)|=|\sin(c_3)|\leq 1$$

kingW3
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  • Does that mean that, if you passed some other function (e.g. a quadratic) through the same three points, you would get something else, perhaps stronger? For example $-\frac{8}{3\pi^2}x^2+\frac{10}{3\pi}x$ goes through the same three points and has a constant second derivation $-\frac{16}{3\pi^2}=0.5403796\ldots$ –  Sep 13 '19 at 10:41
  • @StinkingBishop Yeah, it would be fun to see if we can find $f''(c) =0$, I haven't been able to find any suitable function that passes through those three points. – kingW3 Sep 13 '19 at 16:51
  • You cannot, and the above quadratic should be a counterexample. It's got a constant second derivative. –  Sep 13 '19 at 16:54
  • @StinkingBishop Yeah but we could find a function that has the max value of the second derivative large and the minimum value $0$ on average it would even out but there might be other conditions which forbid it. – kingW3 Sep 13 '19 at 20:48