- #1
Titan97
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Homework Statement
Let ###f## be double differentiable function such that ##|f''(x)|\le 1## for all ##x\in [0,1]##. If f(0)=f(1), then,
A)##|f(x)|>1##
B)##|f(x)|<1##
C)##|f'(x)|>1##
D)##|f'(x)|<1##
Homework Equations
MVT: $$f'(c)=\frac{f(b)-f(a)}{b-a}$$
The Attempt at a Solution
I first tried using integration.
$$-1\le f''(x) \le 1$$
integrating from 0 to x,
$$-x\le f'(x)-f'(0) \le x$$
Again integrating from 0 to x,
$$-\frac{x^2}{2}\le f(x)-f(0)-f'(0)x \le \frac{x^2}{2}$$
But even though I got an inequality for f(x), I could not remove the constants.
Then I applied Rolle theorem for f(x). Since f(0)=f(1), there exists a point (at least one point) ##c## such that f'(c)=0.
There exists a point ##X\in [c,x]## such that
$$f''(X)=\frac{f(x)-f(c)}{x-c}$$
Here, ##x\in [c,1]##, and since 1>x>c, x-c<1. Also, ##|f''(x)|\le 1##.
$$f''(X)=\frac{f(x)-f(c)}{x-c}$$
So, ##\frac{f(x)-f(c)}{x-c}\le 1## and ##f(x)-f(c)\le {x-c}##. Hence I get the answer D. Is this correct?