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 (Show your working)

Answer 1

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

Find the critical number(s)

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

$$3x^2+cx+1=0$$ $$x=\dfrac{-c\pm\sqrt{c^2-12}}{6}$$

We consider $x\in \R$

$$c^2-12\ge0=>c\le-\sqrt{12}\ or\ c\ge\sqrt{12}$$

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 $y=x-x^3.$

$x\not=0, c\le-\sqrt{12}$ or $c\ge\sqrt{12}.$