Chapter 4.3, Problem 59E

a.

To determine

To find point of inflection and local extrema.

Answer to Problem 59E

The point of inflection at $x=-\frac{7}{4}$ and local extrema at $x=-2,x=-\frac{3}{2}$ .

Explanation of Solution

Given:

The given function is $y=4x^3+21x^2+36x-20$ .

Calculation:

There is almost no leeway in the location of the inflection point and the extrema of $f(x)=a{x}^3+b{x}^2+cx+d$ because of the one inflection point occurs at $x=-\frac{b}{3a}$ , and extrema, if any must be located symmetrically about this value of x.

The derivative of the given function is

  $\begin{array}{l}y\text{'}=12x^2+42x+36\\ y\text{'}=6(x+2)(2x+3)\end{array}$

By substituting $y\text{'}=0$ :

  $\begin{array}{l}y\text{'}=6(x+2)(2x+3)\\ 0=6(x+2)(2x+3)\\ x+2=0\\ x=-2\\ 2x+3=0\\ 2x=-3\\ x=-\frac{3}{2}\end{array}$

These are the point of extrema which are symmetric about point of inflection.

The point of inflection is $x=-\frac{b}{3a}$ .

By comparing the equations above $a=4,b=21$ .

  $\begin{array}{l}x=-\frac{21}{3(4)}\\ x=-\frac{7}{4}\end{array}$

b.

To determine

To find point of inflection and local extrema.

Answer to Problem 59E

The point of inflection at $x=1$ and local extrema at $x=0,x=2$ .

Explanation of Solution

Given:

The given function is $y=-2x^3+6x^2-3$ .

Calculation:

There is almost no leeway in the location of the inflection point and the extrema of $f(x)=a{x}^3+b{x}^2+cx+d$ because of the one inflection point occurs at $x=-\frac{b}{3a}$ , and extrema, if any must be located symmetrically about this value of x.

The derivative of the given function is:

  $\begin{array}{l}y\text{'}=-6x^2+12x\\ y\text{'}=6x(-x+2)\end{array}$

By substituting $y\text{'}=0$

  $\begin{array}{l}y\text{'}=6x(-x+2)\\ 0=6x(-x+2)\\ 6x=0\\ x=0\\ -x+2=0\\ -x=-2\\ x=2\end{array}$

These are the point of extrema which are symmetric about point of inflection.

The point of inflection is $x=-\frac{b}{3a}$ .

By comparing the equations above $a-2,b=6$ .

  $\begin{array}{l}x=-\frac{6}{3(-2)}\\ x=1\end{array}$

c.

To determine

To find point of inflection and local extrema.

Answer to Problem 59E

The point of inflection at $x=-\frac{b}{3a}$ .

Explanation of Solution

Given:

The given function is $y=a{x}^3+b{x}^2+cx+d$ .

Calculation:

There is almost no leeway in the location of the inflection point and the extrema of $f(x)=a{x}^3+b{x}^2+cx+d$ because of the one inflection point occurs at $x=-\frac{b}{3a}$ , and extrema, if any must be located symmetrically about this value of x.

The derivative of the given function is $y\text{'}=3a{x}^2+2bx+c$ .

To have local extrema it will occur at the zeros of $y\text{'}$ .

Since, $y\text{'}$ is quadratic and the graph is parabola and any zeros will be symmetric about the vertex which will also be where $f\text{'}\text{'}(x)=0$ .

  $\begin{array}{l}y\text{'}\text{'}=6ax+2b\\ 0=6ax+2b\\ 6ax=-2b\\ 3ax=-b\\ x=-\frac{b}{3a}\end{array}$

The point of inflection is $x=-\frac{b}{3a}$ .