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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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Please Discuss here about this GDB.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Can anyone please tell me which topics do we have to study to solve this question?

GDB # 2 Dated: Jun 24, 16

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

dy/dx = 3x2 -10x +3
3x2 -10x +3 =0 . so we get x=0.33 and x = 3 …. So in the given interval, critical point is where x=0.33
When x=0.33 then f(x) = 3 and x=0 then f(x) = 1 and when x=1 then f(x) = 0
For the given interval, relative maxima occurs at x =0.33 and it is also absolute maxima in given interval … for given interval absolute minimum is at x=1.

mth101-GDB-solution-26June 2016

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Please Discuss here about this GDB.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

kindly koi gdb k main points ko explain karyga? Meney lectures liye hain but yeh critical point or minima maxima nhi smjh aa raha hai.

first derivative ko 0 k equate krn or equation ko quadratic formula k through solve krna ha to ans 0.33 or 3 ata ha. 

ab in values ko f(x) equation men put krny sy functional values aengi or un values ki base pr ap dekhen k jis value sy minimum ans aa rha ha wo absolute minima hogi jaisy is men agr hm 1 use kr k solve krn to minima ata ha or 0.33 sy solve krn to maxima ata ha . yhan pr 3 ki jga 1 isliey use kia q k hmen jo interval dia gya ha wo 0 to 1 ha or 3 is greater than one so 1 is used here

Bakhtawar .. f(x) myn x=0.33 put karney sy answer 3 kaisy aa rha hy? im getting 1.48.. 

ye 3x2 kaha se aya.... sry my math weak a bit so jst asking :)

x^3 ka derivative lia hy.. thats how it became 3x^2 (according to the rule) 

ok got it, thank u :)

math type ma square kase lete ha.....any idea??

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