f β² ( x ) = 6 x 2 + 2 c x + 2 f'(x)=6x^2+2cx+2 f β² ( x ) = 6 x 2 + 2 c x + 2 Find the critical number(s)
f β² ( x ) = 0 = > 6 x 2 + 2 c x + 2 = 0 f'(x)=0=>6x^2+2cx+2=0 f β² ( x ) = 0 => 6 x 2 + 2 c x + 2 = 0
3 x 2 + c x + 1 = 0 3x^2+cx+1=0 3 x 2 + c x + 1 = 0
x = β c Β± c 2 β 12 6 x=\dfrac{-c\pm\sqrt{c^2-12}}{6} x = 6 β c Β± c 2 β 12 β β We consider x β R x\in \R x β R
c 2 β 12 β₯ 0 = > c β€ β 12 o r c β₯ 12 c^2-12\ge0=>c\le-\sqrt{12}\ or\ c\ge\sqrt{12} c 2 β 12 β₯ 0 => c β€ β 12 β or c β₯ 12 β In this case
x =ΜΈ 0 , c x 2 = β 3 x 3 β x x\not=0, cx^2=-3x^3-x x ξ = 0 , c x 2 = β 3 x 3 β x Substitute
y = 2 x 3 β 3 x 3 β x + 2 x y = 2x^3-3x^3-x+2x y = 2 x 3 β 3 x 3 β x + 2 x
y = β x 3 + x y = -x^3+x y = β x 3 + x Then the minimum and maximum points of every curve in the family of polynomials f ( x ) = 2 x 3 + c x 2 + 2 x f(x)=2x^3+cx^2+2x f ( x ) = 2 x 3 + c x 2 + 2 x lie on the curve y = x β x 3 . y=x-x^3. y = x β x 3 .
x =ΜΈ 0 , c β€ β 12 x\not=0, c\le-\sqrt{12} x ξ = 0 , c β€ β 12 β or c β₯ 12 . c\ge\sqrt{12}. c β₯ 12 β .
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