Asked by Ash
Can you please check my work?
Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function.
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 negative zeros?
Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function.
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 negative zeros?
Answers
Answered by
bobpursley
correct
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