fβ²(x)=6x2+2cx+2
f"(x)=12x+2c
fβ²(x)=0
6x2+2cx+2=0
x1β=12β2c+4c2β48ββ
x2β=12β2cβ4c2β48ββ
12x(12β2c+4c2β48ββ)+2c=0
β2c+4c2β48β=β2c
4c2β48=1
c=49/4β=Β±3.5
If c=3.5
fβ²(x)=6x2+7x+2=0
D=49β48=1
x=12β7Β±1ββ
x1β=β0.5
x2β=β2/3
f"(β0.5)=12(β0.5)+2(3.5)=1>0
So, it is minimum
f"(β2/3)=12(β2/3)+2(3.5)=β1<0
So, it is maximum
If c=-3.5
fβ²(x)=6x2β7x+2=0
D=49-48=1
x=127Β±1ββ
x1=0.5
x2=2/3
f"(β0.5)=12(0.5)+2(3.5)=13>0
So it is minimum
f"(2/3)=12(2/3)+2(3.5)=15>0
So it is minimum
In this case
xξ =0,cx2=β3x3βx
Substitute
y=2x3β3x3βx+2x
y=βx3+x
Then the minimum and maximum points of every curve in the family of polynomials f(x)=2x3+cx2+2x lie on the curve y=x-x3
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