$f$ is a function twice differentiable on $[0,1]$ and such that $f''$ is continuous. We suppose that : $f(1)-1=f(0)=f'(1)=f'(0)=0$. Prove that there exists $x_0$ on $[0,1]$ such that $|f''(x_0)| \geq 4$
Problem
Source: Morocco TST 2010, problem 1
Tags: function, algebra, Integral inequality
18.04.2015 13:18
tchebytchev wrote: $f$ is a function twice differentiable on $[0,1]$ and such that $f''$ is continuous. We suppose that : $f(1)-1=f(0)=f'(1)=f'(0)=0$. Prove that there exists $x_0$ on $[0,1]$ such that $|f''(x_0)| \geq 4$ If $f''(x)<4$ $\forall x\in[0,\frac 12]$, then : $f'(x)=\int_0^xf''(t)dt$ $<\int_0^x4dt=4x$ and so $f'(x)<4x$ $\forall x\in[0,\frac 12]$ and then : $f(\frac 12)=\int_0^{\frac 12}f'(t)dt$ $<\int_0^{\frac 12}4tdt=\frac 12$ and so $f(\frac 12)<\frac 12$ If $f''(x)>-4$ $\forall x\in[\frac 12,1]$, then : $f'(x)=-\int_x^1f''(t)dt<\int_x^14dt=4-4x$ and so $f'(x)<4-4x$ $\forall x\in[\frac 12,1]$ and then : $1-f(\frac 12)=\int_{\frac 12}^1f'(t)dt$ $<\int_{\frac 12}^1(4-4t)dt$ $=\frac 12$ and so $f(\frac 12)>\frac 12$ So either $f''(x)\ge 4$ for some $x\in[0,\frac 12]$, either $f''(x)\le-4$ for some $x\in[\frac 12,1]$ And so $|f''(x)|\ge 4$ for some $x\in[0,1]$ Q.E.D.
18.04.2015 13:29
Well done
11.02.2017 13:40
Is there a way without integrals.
12.02.2017 12:37
adityaguharoy wrote: Is there a way without integrals. Here is my solution https://www.dropbox.com/s/gsuo368rcv537mj/My%20solution%20using%20Taylor%27s%20Theorem_1.jpg?dl=0
14.02.2017 08:20
Ani2000 wrote: adityaguharoy wrote: Is there a way without integrals. Here is my solution https://www.dropbox.com/s/gsuo368rcv537mj/My%20solution%20using%20Taylor%27s%20Theorem_1.jpg?dl=0 Your Rolle's theorem argument is wrong
14.02.2017 11:23
AHKR wrote: Ani2000 wrote: adityaguharoy wrote: Is there a way without integrals. Here is my solution https://www.dropbox.com/s/gsuo368rcv537mj/My%20solution%20using%20Taylor%27s%20Theorem_1.jpg?dl=0 Your Rolle's theorem argument is wrong Can you tell me why?
14.02.2017 16:54
okay here is my corrected solution put $x=\frac{1}{2}$ and $x_{0}=1$ we get, $$f\left(\frac{1}{2}\right)=1+\frac{f''(c)}{8}$$where $c\in\left(\frac{1}{2},1\right).$ and, Putting $x=\frac{1}{2}$ and $x_{0}=0,$ we get $$f\left(\frac{1}{2}\right)=\frac{f''(c')}{8}$$where $c'\in\left(0,\frac{1}{2}\right).$ now equating this two equations we get, $$f"(c')-f"(c)=8\implies|f"(c')-f"(c)|=8\implies |f"(c')|+|f"(c)|\ge8$$which implies one of $|f"(c)|,|f"(c')|\ge4$