Chapter 4.4, Problem 35E

(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 35E

The local minima is at $x=\pm \sqrt{\frac{1}{c}},c>0$

Explanation of Solution

Given: $f(x)=c{x}^4-2x^2+1$

Given: $f(x)=c{x}^4-2x^2+1$

  $\begin{array}{l}f(x)=c{x}^4-2x^2+1\\ f\text{'}(x)=4c{x}^3-4x\\ f\text{'}\text{'}(x)=12c{x}^2-4\end{array}$

Finding the solutions for

  $\begin{array}{l}f\text{'}(x)=0\\ 4c{x}^3-4x=0\\ 4x(c{x}^2-1)=0\\ 4x=0\\ x=0\\ c{x}^2-1=0\\ x=\pm \sqrt{\frac{1}{c}}\end{array}$

The graph of the function is

For

  $\begin{array}{l}f\text{'}\text{'}(x)=0\\ 4c{(}^0-4<0\\ f\text{'}\text{'}\left(\pm \sqrt{\frac{1}{c}}\right)=12c{\left(\pm \sqrt{\frac{1}{c}}\right)}^2-4\\ =12c\left(\frac{1}{c}\right)-4\\ =12-4>0\end{array}$

Thus

The local minima is at $x=\pm \sqrt{\frac{1}{c}},c>0$

(b)

To determine

To Show: The maximum and minimum points of every curve in the family lie on the parabola $y=1-x^2$ .

Explanation of Solution

Given: $f(x)=c{x}^4-2x^2+1$

Given: $f(x)=c{x}^4-2x^2+1$

The local minima is at $x=\pm \sqrt{\frac{1}{c}},c>0$

Substituting the value of $x$ we get

  $\begin{array}{l}y=f\left(\pm \sqrt{\frac{1}{c}}\right)\\ y=c{\left(\pm \sqrt{\frac{1}{c}}\right)}^4-2{\left(\pm \sqrt{\frac{1}{c}}\right)}^2+1\\ y=\frac{-1}{c}+1\\ y=-x^2+1\end{array}$

Now,

For $y=-x^2+1$

The graph of the function is

  

The $y$ coordinate of the local minimum is

  $f(0)=c{(0)}^4-2{(0)}^2+1=1$

Thus the local maxima is at $(0,1)$

Showing that the local maxima lies on $y=1-x^2$

Substitute $(0,1)$

Then

  $\begin{array}{l}1=-{(}^0+1\\ 1=1\end{array}$

Hence proved .