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).](https://www.tinkutara.com/question/Q1133.png)
$${Let}\:{f}\::\:\left[\:\mathrm{0}\:,\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}\:\:{be}\:{a}\: \\ $$$${differentiable}\:{function}.\:{Prove} \\ $$$${that}\:{there}\:{exists}\:{a}\:{c}\:\in\:\left[\mathrm{0},\mathrm{1}\right]\:{such} \\ $$$${that}\: \\ $$$$\frac{\mathrm{4}}{\pi}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]=\left(\mathrm{1}+{c}^{\mathrm{2}} \right){f}^{\:} '\left({c}\right).\: \\ $$
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)](https://www.tinkutara.com/question/Q1137.png)
$$\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{diferrentiable}\:\mathrm{into}\:\left[{a},{b}\right]\:\mathrm{then} \\ $$$$\exists\xi\in\left[{a},{b}\right]\Rightarrow{f}'\left(\xi\right)=\frac{{f}\left({b}\right)−{f}\left({a}\right)}{{b}−{a}}\:\:\:\left({b}>{a}\right) \\ $$
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](https://www.tinkutara.com/question/Q1138.png)
$${f}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:\left[{a},{b}\right]\:\mathrm{then} \\ $$$${m}=\mathrm{min}\left\{{f}\left({a}\right),{f}\left({b}\right)\right\} \\ $$$${M}=\mathrm{max}\left\{{f}\left({a}\right),{f}\left({b}\right)\right\} \\ $$$${c}\in\left[{m},\mathrm{M}\right]\Rightarrow\exists\xi\in\left[{a},{b}\right],{f}\left(\xi\right)={c} \\ $$
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](https://www.tinkutara.com/question/Q1141.png)
$${f}'\left({c}\right)=\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi\left(\mathrm{1}+{c}^{\mathrm{2}} \right)} \\ $$$${f}\left({c}\right)=\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi}\mathrm{arctan}\:{c} \\ $$
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))](https://www.tinkutara.com/question/Q1144.png)
$${f}\left({x}\right)={ax}+{b} \\ $$$${f}\left(\mathrm{1}\right)={a}+{b} \\ $$$${f}\left(\mathrm{0}\right)={b} \\ $$$${f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)={a} \\ $$$$\frac{\mathrm{4}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]}{\pi}=\frac{\mathrm{4}{a}}{\pi} \\ $$$${f}'\left({x}\right)={a} \\ $$$$\left(\mathrm{1}+{c}^{\mathrm{2}} \right){f}'\left({c}\right)=\left(\mathrm{1}+{c}^{\mathrm{2}} \right){a} \\ $$$${c}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{4}}{\pi} \\ $$$${c}=\pm\sqrt{\frac{\mathrm{4}}{\pi}−\mathrm{1}} \\ $$
Commented by 123456 last updated on 03/Jul/15
$${g}\left({x}\right)=\mathrm{arctan}\:{x} \\ $$
Answered by 123456 last updated on 03/Jul/15
$${g}\left({x}\right)=\mathrm{arctan}\:{x} \\ $$$$\frac{{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)}{{g}\left(\mathrm{1}\right)−{g}\left(\mathrm{0}\right)}=\frac{{f}'\left({c}\right)}{{g}'\left({c}\right)} \\ $$