133k views
3 votes
How many negative real zeroes does f(x) have if f(-x) has 5 variations in signs

1 Answer

3 votes

Final answer:

Descartes' Rule of Signs implies that if the transformed function f(-x) has 5 changes in signs, the function f(x) may have up to 5 negative real zeros. The exact number cannot be determined without additional details.

Step-by-step explanation:

The question revolves around the number of negative real zeros of a function, given that its transformation, f(-x), has 5 variations in sign. According to Descartes' Rule of Signs, f(-x) having five variations in sign suggests that there could be 5 or less by an even number negative real roots for the function f(x).

However, without more information, such as the degree of the polynomial or additional context, it's not possible to provide the exact number of negative real zeroes that f(x) has. Yet, we can state with certainty that the maximum number of negative real zeros will be 5.

User Neablis
by
8.0k points