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

Algebra ->  Polynomials-and-rational-expressions -> 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      Log On


   



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) About Me  (Show Source):
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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 -6x%5E5+%2B+4x%5E3+-+2x+=+-2x%283x%5E4-2x%5E2+%2B+1%29. Since the factor 3x%5E4-2x%5E2+%2B+1 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:
-x%5E6+%2B+x%5E4+-+x%5E2+%2B+3
graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+-x%5E6+%2B+x%5E4+-+x%5E2+%2B+3+%29