Let f (x) = 2x-1 (x)

Let f (x) = 2x-1 (x)

This is a linear function
There is no maximum value
k>0
So it's an increasing function
D right

The function f (x) = 2x cosx ()
A. Is an increasing function B. is a decreasing function C. has a maximum D. has a minimum

Since the derivative of function f (x) = 2x cosx is f '(x) = 2 + SiNx > 0, the function f (x) = 2x cosx is an increasing function in (- ∞, + ∞), so a

It is known that the minimum value of the image y of the function y = ax quadratic + BX + C is - 3, and the root of the equation AX square + BX + C + 2 = 0 is?

The minimum value of y = ax ^ 2 + BX + C is - 3, the minimum value = > a > 0y = ax ^ 2 + BX + C = a (x + B / 2a) ^ 2 + (4ac-b ^ 2) / 4A (4ac-b ^ 2) / 4A = - 3 = > B ^ 2 = 4A (c + 3) ax ^ 2 + BX + C + 2 = 0 = > ax ^ 2 + BX + C = - 2 > - 3 = > the equation has two solutions, x = [- B + (or -) √ (b ^ 2-4a (c + 2))] / 2A = [- B + (or -) 2 √ a

Find the maximum and minimum of the square X of the function y = 2 / 2 root 3sinxcosx-2 / 2 cos,
And the set of X that causes the function to take these values

Y = (√ 3 / 2) sinxcosx - (1 / 2) cos & # 178; X = (√ 3 / 4) · (2sinxcosx) - (1 / 2) · (1 + cos2x) / 2 = (1 / 2) [sin2x · (√ 3 / 2) - cos2x · (1 / 2)] - 1 / 2 = (1 / 2) sin (2x - π / 3) - 1 / 2. ∧ sin (2x - π / 3) = 1, that is, when x = k π + (5 / 12) π, the maximum value is: 0; sin (2x