Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function. Please check my answers?
1. f(x)=-7x^9+x^5-x^2+6
This is what I did:
There are 2 sign changes
f(-x)=-7(-x)^9+(-x)^5-(-x)^2+6
There are 2 sign changes
So, are there 2 or 0 positive zeros, 2 or 0 negative zeros?
2. f(x)10x^3-8x^2+x+5
There are 2 sign changes
f(-x)=10(-x)^3-8(-x)^2+(-x)+5
There are 2 sign changes
So, are there 2 or 0 positive zeros, no tive zeros?
To use Descartes' Rule of Signs, you need to count the number of sign changes in the coefficients of the polynomial equation. However, you also need to be careful with the signs when substituting -x into the equation.
1. f(x) = -7x^9 + x^5 - x^2 + 6
When determining the signs for f(-x), you made a mistake in the signs of the odd-degree terms. It should be:
f(-x) = -1(-7)(-x)^9 + (-x)^5 - (-x)^2 + 6
Starting with the original equation:
f(x) = -7x^9 + x^5 - x^2 + 6
There is one sign change from -7x^9 to x^5 (positive to negative).
Now let's evaluate f(-x):
f(-x) = -1(-7)(-x)^9 + (-x)^5 - (-x)^2 + 6
f(-x) = -7x^9 - x^5 - x^2 + 6
There are two sign changes from -7x^9 to -x^5 (negative to positive) and from -x^5 to -x^2 (positive to negative).
So, according to Descartes' Rule of Signs for f(x), there are 1 positive zero (the number of sign changes in the original equation) and 2 negative zeros (the number of sign changes in f(-x) when divided by 2 and rounded down).
2. f(x) = 10x^3 - 8x^2 + x + 5
Counting the sign changes in the original equation:
There is one sign change from 10x^3 to -8x^2.
Now evaluating f(-x):
f(-x) = 10(-x)^3 - 8(-x)^2 + (-x) + 5
f(-x) = -10x^3 - 8x^2 - x + 5
There is one sign change from -10x^3 to -8x^2.
According to Descartes' Rule of Signs for f(x), there are 1 positive zero and 0 negative zeros.
So, to summarize:
1. f(x) = -7x^9 + x^5 - x^2 + 6
Possible number of positive zeros: 2 or 0
Possible number of negative zeros: 2 or 0
2. f(x) = 10x^3 - 8x^2 + x + 5
Possible number of positive zeros: 1
Possible number of negative zeros: 0