IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

nandithac

prove that the function h(x)=x^3+x^2+x+1 do not have maxima or minima

arpan1

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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richa_dpsvk

@ 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

soham406

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

arpan1

richa

understand my solution completely and then argue..

try thinking graphically

arpan1

richa

understand my solution completely and then argue..

try thinking graphically

richa_dpsvk

@ 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?

arpan1

not within a subset of real numbers

richa_dpsvk

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

arpan1

that(s) true

man111

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.

arpan1

hey man 111
hav u read the question properly

avi_1214545

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.

avi_1214545

tell me if m wrong..

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