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GDB # 2 | Dated: Jun 24, 16 |
f(x) = {x^3} - 5{x^2} + 3x + 1,
Note:
Its submission time is 30th June 2016 to 4th July 2016. |
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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
Please Discuss here about this GDB.Thanks
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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 3x^{2 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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