Answer to Question #338882 in Calculus for Junior

Question

For what values of 𝑐 does the curve 𝑓(𝑥) = 2𝑥^3 + 𝑐𝑥^2 + 2𝑥 have maximum and minimum points?

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.

Suppose 𝑓 is differentiable on ℝ and has two roots. Show that 𝑓′ has at least one root.

Answer 1

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 $x\in \R$

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

The curve $𝑓(𝑥) = 2𝑥^3 + 𝑐𝑥^2 + 2𝑥$ have maximum and minimum for $c\in(-\infin, -\sqrt{12})\cup (\sqrt{12}, \infin).$

2.

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

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 $c\ge\sqrt{12}.$

3.

We say that $f$ has roots at $a$ and $b, (a<b),$ and since $f$ is continuous and differentiable everywhere as polynomial, it is continuous on $[a, b]$ and differentiable on $(a, b).$ So by the Rolle’s theorem, there is a number $k$ in $(a, b),$ such that $f'(k)=0.$ Therefore, $k$ is a root of $f'(x)=0.$