Question

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

Answer 1

Rolle's theorem:

Answer 2

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

Answer 3

Mean value theorem:

Answer 4

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

Answer 5

To verify the mean value theorem take,
\[\Large\sf{F(x)=f(x)-f(a)-\left(\frac{x-a}{b-a}\right)\left(f(b)-f(a)\right)}\]

Answer 6

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

Answer 7

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

Answer 8

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....

Answer 9

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

Answer 10

I m weakn proving the rules..