anonymous
  • anonymous
how to prove rolle's theorem on any equation ??
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
If f(x) is defined from [a,b] and if f(a)=f(b) , then there exists a point c , where f ' (c) = 0.
apoorvk
  • apoorvk
okay.. akshat you on this?? cause i am in the middle of typing out a long question. one minute plz, coolbird
anonymous
  • anonymous
Yup...I got this covered..

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anonymous
  • anonymous
ya sure apoorv
anonymous
  • anonymous
You understood this coolbird..right?
anonymous
  • anonymous
ya .. but m waiting for apoorv's ans .. coz it contain 6 marks in exam .. so i hv to write something more for get dat marks
apoorvk
  • apoorvk
hmm... so here you coolbird. now rolle's theorem states that suppose there is a continuous function in [a,b], and is differential in (a,b). and its given that f(a)=f(b). then acc. to rolle, at some point 'c' between a and b (atleast one such point) the function's derivative is zero, which basically means that at that point, the slope becomes zero. i ll show you how graphically...|dw:1333021079962:dw| the above example showed 3 values of 'c' where the function's slope beczme zero. but atleast a minimum one 'c' has to be there. how? lets see.. |dw:1333021225448:dw|
apoorvk
  • apoorvk
i hope you are getting a feel of what's happening.... why the derivative has to, has to become zero at some point between a and b.... because it may increase or decrease after 'a', but it must reach a maximum or minimum value value, after which it will recede or increase, to get equal to f(a) at x=b.
anonymous
  • anonymous
Nice example apoorv...
apoorvk
  • apoorvk
and sorry my english has been totally haywire... i typed all that our in a real hurry!
anonymous
  • anonymous
I just wish I could give you two medals.. :P
apoorvk
  • apoorvk
haha thanks.. your words are praise enough. :)
anonymous
  • anonymous
thanX .. @apoorvk n @Akshat4-10 ..
anonymous
  • anonymous
Anytime.. :)

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