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Chapter 2: Problem 94

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

Short Answer

Expert verified
The function \(f(x)=3 x^{3}+2 x^{2}+x+3\) could possibly have 0 positive and 1 or 3 negative real zeros.

Step by step solution

01

- Analysis of the Function

First look at the function \(f(x)=3 x^{3}+2 x^{2}+x+3\). We can see that there are no changes in signs (from positive to negative or vice versa), so according to Descartes' Rule of Signs, there are no positive real zeros.
02

- Analysis of the Reciprocal Function

Next, substitute \(-x\) for \(x\) in the function to get its reciprocal. The reciprocal function is \(f(-x)=3 (-x)^{3}+2 (-x)^{2}+(-x)+3\), which simplifies to \(-3x^{3}+2x^{2}-x+3\). In this case, there are 3 sign changes - from negative to positive between the first and second term, from positive to negative between the second and third term, and from negative to positive between the third and fourth term. Therefore, Descartes' Rule of Signs tells us that there could be 3 or one less than multiple of 2, i.e. 1 negative real zero.
03

- Combine Results

The results show that it's possible for the function \(f(x)=3 x^{3}+2 x^{2}+x+3\) to have 0 positive real zeros and 1 or 3 negative real zeros.

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