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

Let f : [ 0 , 1 ] → R  be a   differentiable function. Prove  that there exists a c ∈ [0,1] such  that   (4/π)[f(1)−f(0)]=(1+c^2 )f^  ′(c).

$${Let}\:{f}\::\:\left[\:\mathrm{0}\:,\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}\:\:{be}\:{a}\: \\ $$$${differentiable}\:{function}.\:{Prove} \\ $$$${that}\:{there}\:{exists}\:{a}\:{c}\:\in\:\left[\mathrm{0},\mathrm{1}\right]\:{such} \\ $$$${that}\: \\ $$$$\frac{\mathrm{4}}{\pi}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]=\left(\mathrm{1}+{c}^{\mathrm{2}} \right){f}^{\:} '\left({c}\right).\: \\ $$

Commented by 123456 last updated on 29/Jun/15

if f is diferrentiable into [a,b] then  ∃ξ∈[a,b]⇒f′(ξ)=((f(b)−f(a))/(b−a))   (b>a)

$$\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{diferrentiable}\:\mathrm{into}\:\left[{a},{b}\right]\:\mathrm{then} \\ $$$$\exists\xi\in\left[{a},{b}\right]\Rightarrow{f}'\left(\xi\right)=\frac{{f}\left({b}\right)−{f}\left({a}\right)}{{b}−{a}}\:\:\:\left({b}>{a}\right) \\ $$

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∈[m,M]⇒∃ξ∈[a,b],f(ξ)=c

$${f}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:\left[{a},{b}\right]\:\mathrm{then} \\ $$$${m}=\mathrm{min}\left\{{f}\left({a}\right),{f}\left({b}\right)\right\} \\ $$$${M}=\mathrm{max}\left\{{f}\left({a}\right),{f}\left({b}\right)\right\} \\ $$$${c}\in\left[{m},\mathrm{M}\right]\Rightarrow\exists\xi\in\left[{a},{b}\right],{f}\left(\xi\right)={c} \\ $$

Commented by 123456 last updated on 30/Jun/15

f′(c)=((4[f(1)−f(0)])/(π(1+c^2 )))  f(c)=((4[f(1)−f(0)])/π)arctan c

$${f}'\left({c}\right)=\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi\left(\mathrm{1}+{c}^{\mathrm{2}} \right)} \\ $$$${f}\left({c}\right)=\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi}\mathrm{arctan}\:{c} \\ $$

Commented by 123456 last updated on 01/Jul/15

f(x)=ax+b  f(1)=a+b  f(0)=b  f(1)−f(0)=a  ((4[f(1)−f(0)])/π)=((4a)/π)  f′(x)=a  (1+c^2 )f′(c)=(1+c^2 )a  c^2 +1=(4/π)  c=±(√((4/π)−1))

$${f}\left({x}\right)={ax}+{b} \\ $$$${f}\left(\mathrm{1}\right)={a}+{b} \\ $$$${f}\left(\mathrm{0}\right)={b} \\ $$$${f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)={a} \\ $$$$\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi}=\frac{\mathrm{4}{a}}{\pi} \\ $$$${f}'\left({x}\right)={a} \\ $$$$\left(\mathrm{1}+{c}^{\mathrm{2}} \right){f}'\left({c}\right)=\left(\mathrm{1}+{c}^{\mathrm{2}} \right){a} \\ $$$${c}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{4}}{\pi} \\ $$$${c}=\pm\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}} \\ $$

Commented by 123456 last updated on 03/Jul/15

g(x)=arctan x

$${g}\left({x}\right)=\mathrm{arctan}\:{x} \\ $$

Answered by 123456 last updated on 03/Jul/15

g(x)=arctan x  ((f(1)−f(0))/(g(1)−g(0)))=((f′(c))/(g′(c)))

$${g}\left({x}\right)=\mathrm{arctan}\:{x} \\ $$$$\frac{{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)}{{g}\left(\mathrm{1}\right)−{g}\left(\mathrm{0}\right)}=\frac{{f}'\left({c}\right)}{{g}'\left({c}\right)} \\ $$

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