Chapter 4.4, Problem 36E

(a)

To determine

To Describe:The family of polynomials given by the equation. For what values of $c$ the curve has minimal point.

Answer to Problem 36E

The local minima is at $c\ge \sqrt{12},c\ge -\sqrt{12}$

Explanation of Solution

Given: $f(x)=2x^3+c{x}^2+2x$

Given: $f(x)=2x^3+c{x}^2+2x$

  $\begin{array}{l}f(x)=2x^3+c{x}^2+2x\\ f\text{'}(x)=6x^2+2cx+2\end{array}$

Finding the solutions for $f\text{'}(x)$

  $\begin{array}{l}3x^2+cx+1=0\\ x=\sqrt{c-12}\end{array}$

For

  $\begin{array}{l}{c}^2\ge 12\\ c\ge \sqrt{12},c\ge -\sqrt{12}\end{array}$

Thus

The minimal point is at $c\ge \sqrt{12},c\ge -\sqrt{12}$

(b)

To determine

To Show: The maximum and minimum points of every curve in the family lie on the parabola $y=x-x^3$ .

Explanation of Solution

Given: $f(x)=2x^3+c{x}^2+2x$

Given: $f(x)=2x^3+c{x}^2+2x$

The minimal point is at $c\ge \sqrt{12},c\ge -\sqrt{12}$

Substituting the value of $x$ we get

  $\begin{array}{l}f\text{'}(x_0)=0\\ 3x_0{}^2+c{x}_0+1=0\\ c=-\frac{1+3x_0{}^2}{x_0{}^2}\end{array}$

Now,

For $\begin{array}{l}f(x_0)=x_0-x_0{}^3\\ (x_0,y_o)=(x_0,x_0-x_0{}^3)\end{array}$

Which means that the lies on the curve $y=x-x^3$

The graph of the function is

  

Hence proved.