explain: f(x)= 3x^2, F(x) = ax^3 Solution 5.1.16 Let f(x) = ax3 + bx2 + cx + d,

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explain: f(x)= 3x^2, F(x) = ax^3

Explanation / Answer

5.1.16 Let f(x) = ax3 + bx2 + cx + d, with a, b, c, d real numbers. a?0. Then f '(x) = 3ax2 + 2bx + c. Since f'(x) is a quadratic polynomial (degree 2), it can have at most two real roots. Therefore f(x) can have at most 2 critical points. If f(x) = x3 - 6x, then f '(x) = 3x2 - 6, which has two distinct real roots. Therefore f(x) has two critical points. If f(x) = x3, then f '(x) = 3x2, which has one real root. Therefore f(x) has one critical point. If f(x) = x3 + 6, then f '(x) = 3x2 + 6, which has no real roots. Therefore f(x) has no critical points. 5.1.17 Let f(x) = x3 + cx +1. Then f '(x) = 3x2 + c. If c > 0, then f '(x) has no real roots, and since f ' is continous, this implies that f(x) has no local extrema. If c = 0, then f '(x) has one real root at x = 0. However, f '(x) > 0 for all x, so f is everywhere increasing, and therefore f(x) has no local extrema. If c0, so that x = --4*3*c6 is a local minimum, and x = +-4*3*c6 is a local maximum. 5.1.18 Let f(x) be a polynomial of degree n. Then f(x) = anxn + an-1xn-1 + … + a1x + a0, with an ? 0. Then f '(x) = nanxn-1 + (n-1)an-1xn-2 + … + a2x + a1. By the fundamental theorem of algebra, since f '(x) is an (n-1) degree polynomial, it can have at most (n-1) roots. Since f ' is continuous on the reals, these are the only critical points of f(x). Therefore f has at most (n-1) critical points.

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