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

1. amilapsn

Rolle's theorem:

2. amilapsn

$$(a<b)\wedge (f~is~continuous~on[a,b])\\\wedge (f~is~differentiable~on(a,b))\wedge (f(a)=f(b)=0)\\ \Rightarrow \exists c\in(a,b)~s.t.f^/(c)=0$$

3. amilapsn

Mean value theorem:

4. amilapsn

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

5. amilapsn

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

and apply Rolle's theorem to $$F(x)$$

7. anonymous

i cant understaunderstand the verification.. can you send me a link of that.. plz

8. amilapsn

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

Oh okay dear.. from where i cn gwt that derivatuins?? :(

10. anonymous

I m weakn proving the rules..

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