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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 12:41:29 IST
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prove that the function h(x)=x^3+x^2+x+1 do not have maxima or minima
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 12:50:02 IST
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this function has 2 complex and 1 real root... therefore there is no point in comparision of maxima or minima within a subset of real numbers
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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 plz rate if u like my ans
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 12:58:49 IST
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richa
understand my solution completely and then argue..
try thinking graphically
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:00:18 IST
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richa
understand my solution completely and then argue..
try thinking graphically
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:03:27 IST
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@ arpan
Look, suppose we have a cubic with only one real root and two complex roots....do you mean to say that it wont have extreme points?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:05:19 IST
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not within a subset of real numbers
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:08:09 IST
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you are confused dude!! :)
what have extreme points got to do with the roots being real....tell me, the quadratic expression x^2 + x + 1 has no real root right?
still, it has a minima at x = -1/2, the value of the minima being 3/4
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:17:54 IST
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that(s) true
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 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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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:33:21 IST
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hey man 111 hav u read the question properly
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:41:17 IST
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h(x)=x3+x2+x+1 differentiating, we get h'(x)=3x2+2x+1 now, for the function, h'(x) D=22 - 4*1*3 = -8 hence, a is +ive and D is -ive... this implies that the function, h'(x) is +ive x R
hence, nowhere, on real axis, h'(x) is 0
consequently..no extrema.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:45:39 IST
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tell me if m wrong..
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Na pine ka shouk tha na pilane ka shonk tha. Hume to sirf NAZREIN milane ka shonk tha. Par kya karen hum nazre hi unse mila baithe jinhe nazron se PILANE ka shonk tha. |
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