Chapter 9.1, Problem 60E

To determine

To find: the number of possible positive and negative zeros.

Answer to Problem 60E

The number of positive real zeros are either three or one and the negative zeros are zero.

Explanation of Solution

Given information:

The function is $f\left(x\right)=x^3-4x^2+4x-1$ .

Calculation: to find the number of positive real zeros, find the change in sign,

  $f\left(x\right)=x^3-4x^2+4x-1$

In the function sign change three times so the number of positive zeros are either three or one.

Now substitute $x=-x$ in the function,

  $\begin{array}{l}f\left(x\right)=x^3-4x^2+4x-1\\ f\left(-x\right)={\left(-x\right)}^3-4{\left(-x\right)}^2+4\left(-x\right)-1\text{ [susbtitute]}\\ f\left(-x\right)=-x^3-4x^2-4x-1\text{ [simplify]}\end{array}$

In the function the sign is unchanged,

Thus the number of negative zeros is zero.