Let-f-0-1-R-be-a-differentiable-function-Prove-that-there-exists-a-c-0-1-such-that-4-pi-f-1-f-0-1-c-2-f-c-

Question Number 1133 by 112358 last updated on 29/Jun/15

L e t f : [0, 1] \to R b e a

d i f f e r e n t i a b l e f u n c t i o n . P r o v e

t h a t t h e r e e x i s t s a c \in [0, 1] s u c h

t h a t

Prime causes double exponent: use braces to clarify

Commented by 123456 last updated on 29/Jun/15

if f is diferrentiable into [a, b] then

\exists ξ \in [a, b] \Rightarrow f^{'} (ξ) = \frac{f (b) - f (a)}{b - a} (b > a)

Commented by 123456 last updated on 29/Jun/15

f is continuous at [a, b] then

m = \min {f (a), f (b)}

M = \max {f (a), f (b)}

c \in [m, M] \Rightarrow \exists ξ \in [a, b], f (ξ) = c

Commented by 123456 last updated on 30/Jun/15

f^{'} (c) = \frac{4 [f (1) - f (0)]}{π (1 + c^{2})}

f (c) = \frac{4 [f (1) - f (0)]}{π} \arctan c

Commented by 123456 last updated on 01/Jul/15

f (x) = a x + b

f (1) = a + b

f (0) = b

f (1) - f (0) = a

\frac{4 [f (1) - f (0)]}{π} = \frac{4 a}{π}

f^{'} (x) = a

(1 + c^{2}) f^{'} (c) = (1 + c^{2}) a

c^{2} + 1 = \frac{4}{π}

c = \pm \sqrt{\frac{4}{π} - 1}

Commented by 123456 last updated on 03/Jul/15

g (x) = \arctan x

Answered by 123456 last updated on 03/Jul/15

g (x) = \arctan x

\frac{f (1) - f (0)}{g (1) - g (0)} = \frac{f^{'} (c)}{g^{'} (c)}

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