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anonymous

  • one year ago

State and verify rolle's theorem and state and verify mean value theorem..

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  1. amilapsn
    • one year ago
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    Rolle's theorem:

  2. amilapsn
    • one year ago
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    \((a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\wedge (f(a)=f(b)=0)\\ \Rightarrow \exists c\in(a,b)~s.t.f^/(c)=0\)

  3. amilapsn
    • one year ago
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    Mean value theorem:

  4. amilapsn
    • one year ago
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    \((a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\\ \Rightarrow \exists c\in(a,b)~s.t.f^/(c)=\frac{f(b)-f(a)}{b-a}\)

  5. amilapsn
    • one year ago
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    To verify the mean value theorem take, \[\Large\sf{F(x)=f(x)-f(a)-\left(\frac{x-a}{b-a}\right)\left(f(b)-f(a)\right)}\]

  6. amilapsn
    • one year ago
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    and apply Rolle's theorem to \(F(x)\)

  7. anonymous
    • one year ago
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    i cant understaunderstand the verification.. can you send me a link of that.. plz

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

  9. anonymous
    • one year ago
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    Oh okay dear.. from where i cn gwt that derivatuins?? :(

  10. anonymous
    • one year ago
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    I m weakn proving the rules..

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