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Chapter 2: Problem 92

Use Descartes's Rule of Signs to determine the possible numbers of positive and negative real zeros of the function. $$f(x)=4 x^{3}-3 x^{2}+2 x-1$$

Short Answer

Expert verified
By using Descartes's Rule of Signs, it can be seen that the given function \(f(x)=4 x^{3}-3 x^{2}+2 x-1\) possibly has 2 or 0 positive real roots and no negative real roots.

Step by step solution

01

Identify the Coefficients

The given function is \(f(x)=4 x^{3}-3 x^{2}+2 x-1\). The coefficients of this polynomial are 4, -3, 2, -1.
02

Apply Descartes's Rule of Signs for Positive Roots

To find the possible number of positive real roots, we check the sign changes in the coefficients. The signs are +, -, +, -. So, the sign changes twice. Therefore, the function has 2 or 0 positive real roots.
03

Apply Descartes's Rule of Signs for Negative Roots

To find the possible number of negative real roots, we replace each \(x\) by \(-x\) in the function which yields \(f(-x) = 4(-x)^3 - 3(-x)^2 + 2(-x) - 1 = -4x^3 - 3x^2 -2x -1\). The signs of the coefficients are now -, -, -, -. There are no sign changes, hence, the function has no negative real roots.

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