Chapter 6.7, Problem 59HP

(a)

To determine

To give: The counterexample for the given statement.

Answer to Problem 59HP

The counter example is explained below.

Explanation of Solution

Given:

All polynomial functions of degree greater than 2 have at least 1 negative real root.

Calculation:

Consider the function.

  $f\left(x\right)=x^3+4x^2+64$

Find the total number o zeros of the above function.

There are no sign changes for the coefficients of $f\left(x\right)$ , hence the function has 0 positive real zeros.

  $f\left(-x\right)=-x^3+4x^2+64$

There is 1 sign change for the coefficients of $f\left(-x\right)$ , so $f\left(x\right)$ has 1 negative real zeros.

(b)

To determine

To give: The counterexample for the given statement.

Answer to Problem 59HP

The counter example is explained below.

Explanation of Solution

Given:

All polynomial functions of degree greater than 2 have at least 1 positive real root.

Calculation:

Consider the function.

  $f\left(x\right)=x^3+6x^2+9x$

Find the total number of zeros of the above function.

There are no sign changes for the coefficients of $f\left(x\right)$ , hence the function has 3 positive real zeros.

  $f\left(-x\right)=-x^3+6x^2-9x$

There is 2 sign change for the coefficients of $f\left(-x\right)$ , so $f\left(x\right)$ has 1 or 2 negative real zeros.