Chapter 5.3, Problem 61E

a.

To determine

To check the statement by examining for cubic with equation $y=4x^3+21x^2+36x-20$ .

Explanation of Solution

Given information:

The given statement is that there is almost no leeway in the locations of the inflection point and extrema of $f\left(x\right)=a{x}^3+b{x}^2+cx+d,a\ne 0$ , because the one inflection point occurs at $x=\frac{-b}{3a}$ and extrema, if any, must be located symmetrically about this value of x .

Proof:

The graph of the cubic $y=4x^3+21x^2+36x-20$

  

The inflection point of the graph is at $x=-1.75$ and its extrema are at $x=-2$ and $x=-1.5$ . The extrema are symmetric about the inflection pint since they are both 0.25 unit from the inflection point.

b.

To determine

To check the statement by examining for cubic with equation $y=-2x^3+6x^2-20$ .

Explanation of Solution

Given information:

The given statement is that there is almost no leeway in the locations of the inflection point and extrema of $f\left(x\right)=a{x}^3+b{x}^2+cx+d,a\ne 0$ , because the one inflection point occurs at $x=\frac{-b}{3a}$ and extrema, if any, must be located symmetrically about this value of x .

Proof:

The graph of the cubic $y=-2x^3+6x^2-20$

  

The inflection point of the graph is at $x=1$ and its extrema are at $x=2$ and $x=0$ . The extrema are symmetric about the inflection pint since they are both 1 unit from the inflection point.

c.

To determine

To prove the statement for the general case.

Explanation of Solution

Given information:

The given statement is that there is almost no leeway in the locations of the inflection point and extrema of $f\left(x\right)=a{x}^3+b{x}^2+cx+d,a\ne 0$ , because the one inflection point occurs at $x=\frac{-b}{3a}$ and extrema, if any, must be located symmetrically about this value of x .

Proof:

Take 1^st and 2^nd derivatives

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3a{x}^2+2bx+c\\ f^{″}\left(x\right)=6ax+2b\end{array}$

Apply quadratic formula to get extrema point for which $f^{\prime }\left(x\right)=0$

  $\begin{array}{l}3a{x}^2+2bx+c=0\\ x=\frac{-2b\pm \sqrt{4b^2-12ac}}{6a}\end{array}$

At an inflection point $f^{″}\left(x\right)$ changes sign, which means you need to find points for which $f^{″}\left(x\right)=0$

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

Hence, $-\frac{b}{3a}$ is an unique inflection point.

If $b^2-3ac\ge 0$ there are two extremum point positioned symmetrically around the inflection point at a distance of $\frac{\sqrt{b^2-3ac}}{3a}$