Chapter 4.3, Problem 60E

a.

To determine

To find: To find the point of inflection.

Answer to Problem 60E

The point of inflection at $f\text{'}\text{'}(x)=12x^2+6x+2$ .

Explanation of Solution

Given:

The given function is $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$ with $a\ne 0$ .

Calculation:

The point where the graph of a function has a tangent line and where the concavity change is a point of inflection.

The given function is $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$ .

The derivative of the function is $f\text{'}(x)=4a{x}^3+3b{x}^2+2cx+d$ .

The second derivative of the function is $f\text{'}\text{'}(x)=12a{x}^2+6bx+2c$ .

Since $f\text{'}\text{'}(x)$ is quadratic, it must have $0,1$ or $2$ Zeros. A quadratic with $0$ or $1$ Zeros never change sign, so $f$ has no point of inflection.

If $f\text{'}\text{'}(x)$ has $2$ Zeros, it will change sign twice, and $f$ will have $2$ points of inflection in the diagram of $f$ .

Let $a=b=c=d=e=1$ .

  $f(x)=x^4+x^3+x^2+x+1$

  

The point of inflection is $f\text{'}\text{'}(x)=12x^2+6x+2$ .

Test value for

  $\begin{array}{l}x=0\\ f\text{'}\text{'}(x)=2.00\\ \end{array}$

Sign of $f\text{'}\text{'}$ is +.

b.

To determine

To find: To find condition of point of inflection.

Answer to Problem 60E

The condition for the point of inflection is $b^2>\frac{8}{3}ac$ .

Explanation of Solution

Given:

The given function is $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$ with $a\ne 0$ .

Calculation:

The discriminant of $f\text{'}\text{'}(x)=12x^2+6x+2$ is greater than zero.

A condition that must be satisfied by the coefficient of the graph of $f$ has $0$ or $2$ points of inflection.

  $\begin{array}{c}$ {(6b)}^2-4(12a)(2c)>0$36b^2-96ac>0\\ 12(3b^2-8ac)>0\\ b^2>\frac{8}{3}ac\end{array}$