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Differential Calculus
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19 Mar 2008 12:55:45 IST
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@ Arpan1
ur solution is wrong....the maximum and minima existence does not depend on the nature of roots.....for eg x^3 - 6x^2 + 11x + 100 will have only one real root but it still has two extrema...
to prove that there are no extrema, take its derivative which will be 3x^2 + 2x + 1...this polynomial has no real roots...hence, no extrema
this is the correct proof
ur solution is wrong....the maximum and minima existence does not depend on the nature of roots.....for eg x^3 - 6x^2 + 11x + 100 will have only one real root but it still has two extrema...
to prove that there are no extrema, take its derivative which will be 3x^2 + 2x + 1...this polynomial has no real roots...hence, no extrema
this is the correct proof
19 Mar 2008 12:57:59 IST
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Differntiate the function you get
3x^2 +2x+1
find its discriminant . its always +ve
so in the range of R its strictle increasing / therefore it will have no particular min or max which is actually at + infinity and - infinity
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19 Mar 2008 13:31:18 IST
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here we can write x3+x2+x+1 as (x4-1)/x-1
so f(x)=x4-1/x-1=
f'(x) = (x-1)4x3 - x4/(x-1)2 = x3(4x-4-x)/(x-1)2=
x3(3x-4)/(x-1)2
again calculate {(x-1)2(16x3-12x2) -(4x4-4x3)2(x-1)}/(x-1)4
put f'(x) =0 we get x=0,3/4
for x=0
f''(x) = 0
and x=3/4
we get f''(x) =
find this
after finding f''(x) for x=3/4 u can easily find out the answer.






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