What is the maximum number of negative real roots for the function f(x) = x^5 – 2x^3 + 7x^2 – 2x + 2?

Understanding the Behavior of the Polynomial

The function in question is a polynomial of degree five, which means it can have up to five real roots in total.
One important aspect to consider here is how many of these roots can be negative real roots.

Applying Descartes’ Rule of Signs

To determine the possible number of negative real roots, we can utilize Descartes’ Rule of Signs, which analyzes the sign changes in the polynomial when we substitute -x for x.

Let’s perform this substitution:

f(-x) = (-x)^5 - 2(-x)^3 + 7(-x)^2 - 2(-x) + 2 = -x^5 + 2x^3 + 7x^2 + 2x + 2

Sign Changes in f(-x)

Next, we’ll determine the signs of the terms in the polynomial f(-x) = -x^5 + 2x^3 + 7x^2 + 2x + 2:

  • -x^5 (negative)
  • +2x^3 (positive)
  • +7x^2 (positive)
  • +2x (positive)
  • +2 (positive)

From the sequence of signs, we see that there is one sign change (from negative to positive at -x^52x^3). This indicates that there is exactly one negative real root for the function.

Conclusion

In conclusion, based on an analysis using Descartes’ Rule of Signs, the function f(x) = x^5 - 2x^3 + 7x^2 - 2x + 2 has exactly one negative real root.

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