Continuous function

ABM · Poincare Conjecture Offline Joined: 11 Oct 2005 Posts: 126

need help with the following... Thx

f :[0,1]-> [0,1] a continous function. suppose that f ' exists and conts. on [0,1]
how do u show that there is some c in [0,1] such that l f'(c)l < and =1
and also show that this inequality cannot be improved to l f'(c) l< and = A
for A < 1

what if that f' and f" exist and conts on [0,1] then how to show that there is a c in [0,1] such that l f"(c) l < and = 8 and also show that this inequality cannot be improved to l f"(c) l < and = A for any A < 8

**Posted:** Tue Oct 11, 2005 2:22 pm

Diogene

1°)Remember : $\exists c\in ]a,b[\ s.t.\ f(b)-f(a) = (b-a) f'(c)$ , so $|f'(c)|=|f(1)-f(0)| \leq 1$
Particular case : $f(x)=x \Longrightarrow f'(x)=1$
2°)Remember : $\exists c \ s.t\ f(b) = f(a) + (b-a)f'(a) + \frac{(b-a)^2}2 f''(c)$
Cases to study :
* 2.1) $f'$ has a zero , $f'(a)=0$ .You choose $b=0$ or $b=1$ in order to have $|b-a| \geq \frac 12$ , and you obtain : $|f''(c)| = \frac{2|f(b)-f(a)|}{(b-a)^2} \leq 8$
* 2.2) $f'$ has no zero, $\forall x\ f'(x) > 0$ , so $f$ is increasing . From 1°, there is $a$ such that $0 < f'(a) \leq 1$
** 2.2.1) $a \geq \frac12$ . Take $b=0$ , so $f''(c) = 2\frac{(f(0)-f(a))+af'(a)}{a^2} \Longrightarrow |f''(c)| \leq 8$
** 2.2.2) $a \leq \frac12$ . Take $b=1$ , so $f''(c) = 2\frac{(f(1)-f(a))-(1-a)f'(a)}{(1-a)^2} \Longrightarrow |f''(c)| \leq 8$
* 2.3) $f'$ has no zero, $\forall x\ f'(x) < 0$ . Take $g(x) = 1-f(x)$ so $g''x) = -f''(x) \ > 0$ and from 2.2) there is $c$ s.t. $|f''(c)|=|g''(c)| \leq 8$
Particular case $f(x)= 4x^2-4x +1 \Longrightarrow f''(x)=8$
Cool

**Posted:** Thu Oct 13, 2005 4:19 pm

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