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anonymous
 one year ago
State and verify rolle's theorem and state and verify mean value theorem..
anonymous
 one year ago
State and verify rolle's theorem and state and verify mean value theorem..

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amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1\((a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\wedge (f(a)=f(b)=0)\\ \Rightarrow \exists c\in(a,b)~s.t.f^/(c)=0\)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1\((a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\\ \Rightarrow \exists c\in(a,b)~s.t.f^/(c)=\frac{f(b)f(a)}{ba}\)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1To verify the mean value theorem take, \[\Large\sf{F(x)=f(x)f(a)\left(\frac{xa}{ba}\right)\left(f(b)f(a)\right)}\]

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1and apply Rolle's theorem to \(F(x)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i cant understaunderstand the verification.. can you send me a link of that.. plz

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.1Assume: \[\Large\sf(a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\] take \[\Large\sf{F(x)=f(x)f(a)\left(\frac{xa}{ba}\right)\left(f(b)f(a)\right)}\] \[F(a)=F(b)=0\\ \therefore\text{by rolle's theorem}\\ \exists c\in(a,b)s.t. F^/(c)=0\] by that you can get the mean value theorem..... It's a matter of taking derivatives and simplifications....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay dear.. from where i cn gwt that derivatuins?? :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I m weakn proving the rules..
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