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)$

Question

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.

Answer 1

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 $$|f''(c_3)|=|\sin(c_3)|\leq 1$$