It is known that the functions y = ax and y = - BX are decreasing functions in the interval (0, + ∞). Try to determine the monotone interval of the function y = AX3 + bx2 + 5

It is known that the functions y = ax and y = - BX are decreasing functions in the interval (0, + ∞). Try to determine the monotone interval of the function y = AX3 + bx2 + 5

∵ functions y = ax and y = - BX are both decreasing functions in the interval (0, + ∞), a < 0, B < 0. From y = AX3 + bx2 + 5, we get y ′ = 3ax2 + 2bx. Let y ′ > 0, that is, 3ax2 + 2bx > 0, ■ - 2b3a < x < 0. Therefore, when x ∈ (- 2b3a, 0), the function is an increasing function; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Let y ′ < 0, i.e. 3ax2 + 2bx < 0, X ＜ - 2b3a or x > 0. Therefore, when x ∈ (- ∞, - 2b3a) and (0, + ∞), the function is a decreasing function; the monotone increasing interval of the function y = AX3 + bx2 + 5 is (- 2b3a, 0); the monotone decreasing interval is (- ∞, - 2b3a) and (0, + ∞)

(1) if a is a single element set, find the value of a and set a (2) find the set P = {a belongs to R | a such that a to
(1 / 2) the known set a = {x belongs to R | ax ^ - 3x + 2 = 0}
(1) If a is a single element set, find the value of a and the set a
(2) Find the set P = {a belongs to R | a so that a contains at least

1. If set a is a single element set, then a = 0 or discriminant = 9-8a = 0, then a = 9 / 8;
2、?

Given the set a = {x | ax & sup2; + 3x + 2 = 0, a ∈ r} if there is only one element in a, find the value of a and write this element

If there is only one element, the equation has only one solution
A = 0, one variable linear equation, obviously satisfies
a≠0
Then the discriminant is equal to 0
9-8a=0
a=9/8
Substituting to solve the equation
therefore
a=0,x=-2/3
a=9/8,x=-4/3

Given the set a = {ax square - 3x + 2 = 0, a ∈ r}, if a has only one element, find the value of a and write it out; if there is at most one element% C in a

... a = 9 / 8 elements 4 / 3