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GDB # 2 Dated: Jun 24, 16
f(x) = {x^3} - 5{x^2} + 3x + 1,
  1. Find its first order derivative. 
  2. Find its critical points in the interval [0,1].  
  3. Find its functional values at critical points and at end points of the interval [0,1].
  4. Determine whether its absolute maximum and absolute minimum values occur at critical points or at the end points of the interval [0,1].

 Note:

  • If you successfully able to calculate the first order derivative then you deserve to be 25% marks.
  • If you successfully able to find the critical points then you deserve to be 50% marks.
  • If you further succeeded to calculate the functional values at mentioned points then you deserve to be 75% marks.
  • Finally if you are able to answer the option “d” then you deserve to be 100% marks.

Its submission time is 30th June 2016 to 4th July 2016.

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Replies to This Discussion

Top left par sarey symbols diye huwe hyn.

see

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ahan, ab smjh aya hai... thanks dear for clearing my doubts :)

Dear Students Take Ideas from each other and solve yourself.......Assignments are not for getting solutions only....they are used for your study to boost up your thinking capabilities....If you present a self made Solution ....even if it is wrong....u will get some marks for attempt.....

first derivative leny k bad ye equation ati ha  3x^2 -10x +3

ap ny is pr quadratic formula apply krna ha

3x^2 -10x +3=0

to is k roots 0.33 or 3 aty hn ap check kr len

x ki value 0.33 put krne se f(x) ki value 1.48 aati hy. ye 3 kaise aai hy ??

you are right zainab.... 3 kis trhan aa rha hai yeh nhi smjh aa rahi 

ye dekh len aisy solve krna ha

img sahee ni ha itna pr quadratic formula apply krna ha first derivate pr

sahih hai aa gaya hai smjh me 

but 0.33 ko function me put kr k answer 3 nhi aa rha hai 1.48 aa rha hai 

woh b smjha den 

When x=0.33 then f(x) = 3 and x=0 then f(x) = 1 and when x=1 then f(x) = 0

yeh dekhen ... jo solution provided hai us me yeh likha hai ... 

1.48 he aye ga 3 ni ata

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