The function f (x) = X3 + AX2 + BX + C, where a, B and C are real numbers. When a2-3b < 0, f (x) is () A. Increasing function B. decreasing function C. constant D. is neither increasing function nor decreasing function

The function f (x) = X3 + AX2 + BX + C, where a, B and C are real numbers. When a2-3b < 0, f (x) is () A. Increasing function B. decreasing function C. constant D. is neither increasing function nor decreasing function


If f ′ (x) = 3x2 + 2aX + B, where △ = 4a2-12b < 0, f ′ (x) > 0, then f (x) is an increasing function



It is known that there are two points a (x1, Y1) and B (X2, Y2) on the image with positive scale function y = KX. When x1y2, find the value of K


y1-y2=kx1-kx2=k(x1-x2)>0
Because x1-x2



As shown in the figure, the image of the linear function Y1 = KX + B and the image of the inverse scale function y2 = m / X intersect at two points a (2,1), B (1, - 2)?


The coordinates of B are (- 1, - 2) bar? 1, because a (2,1), B (- 1, - 2) are the points on Y1 = k x + B, then there are: 2K + B = 1, and - K + B = - 2.. solve this equation, get k = 1, B = - 1, so Y1 = X-1.. because a, B are the points on y2 = m / x, so m = 2, that is, y2 = 2 / X.2, from the figure, when - 1 < x < 0 or x > 2, Y1 > Y2



As shown in the figure, it is known that the image of the first-order function Y1 = KX + B and the inverse scale function y2 = MX intersect at points a (- 4, - 2) and B (2,4). (1) find the analytic expressions of these two functions; (2) answer according to the image, when x takes what value, Y1 > Y2?


(1) Substituting points a (- 4, - 2) and B (2,4) into Y1 = KX + B to get: - 4K + B = - 22K + B = 4, the solution is k = 1b = 2. Then the analytic formula of the first-order function is: Y1 = x + 2. Substituting B (2,4) into y2 = MX to get: M = 8, then the analytic formula of the inverse proportional function is: y2 = 8x; (2) when - 4 < x < 0 or x > 2, Y1 > Y2