SOLUTION: Find the maximum number of x-intercepts and the maximum number of turning points that the graph of the function can have. f(x) = � x2 + x4 � x6 + 3

Question 472974: Find the maximum number of x-intercepts and the maximum number of turning points that the graph of the function can have.
f(x) = � x2 + x4 � x6 + 3
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Applying Descartes' rule of signs, the following are the possibilities for the distribution of roots (3 variations of sign):
(i) 3 positive, 3 negative roots, 0 complex
(ii) 3 positive, 1 negative roots, 2 complex
(iii) 1 positive, 3 negative roots, 2 complex
(iv) 1 positive, 1 negative roots, 4 complex
Since the polynomial is symmetric, cases (ii) and (iii) are eliminated.
Taking the derivative, we get . Since the factor does not have real roots, there is only 1 turning point, and it happens at x = 0. This means that there are 2 x-intercepts.
We verify graphically:

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