Question

Show that the minimum and maximum points of every curve in the family of polynomials

𝑓(𝑥) = 2𝑥^3 + 𝑐𝑥^2 + 2𝑥 lie on the curve 𝑦 = 𝑥 − 𝑥^3

Answer 1

$f'(x)=6x^2+2cx+2$

$f"(x)=12x+2c$

$f'(x)=0$

$6x^2+2cx+2=0$

$x_1=\frac{-2c+\sqrt{4c^2-48}}{12}$

$x_2=\frac{-2c-\sqrt{4c^2-48}}{12}$

$12x(\frac{-2c+\sqrt{4c^2-48}}{12})+2c=0$

$-2c+\sqrt{4c^2-48}=-2c$

$4c^2-48=1$

$c=\sqrt{49/4}=\plusmn 3.5$

If c=3.5

$f'(x)=6x^2+7x+2=0$

$D=49-48=1$

$x=\frac{-7 \plusmn \sqrt1}{12}$

$x_1=-0.5$

$x_2=-2/3$

$f"(-0.5)=12(-0.5)+2(3.5)=1>0$

So, it is minimum

$f"(-2/3)=12(-2/3)+2(3.5)=-1<0$

So, it is maximum

If c=-3.5

$f'(x)=6x^2-7x+2=0$

D=49-48=1

$x=\frac{7 \plusmn \sqrt1}{12}$

x₁=0.5

x₂=2/3

$f"(-0.5)=12(0.5)+2(3.5)=13>0$

So it is minimum

f"(2/3)=12(2/3)+2(3.5)=15>0

So it is minimum

In this case

$x\not=0, cx^2=-3x^3-x$

Substitute

$y=2x^3-3x^3-x+2x$

$y=-x^3+x$

Then the minimum and maximum points of every curve in the family of polynomials $f(x)=2x^3+cx^2+2x$ lie on the curve y=x-x³