If x belongs to (0,2) and f (x) = 3x + 1, find f (11)

If x belongs to (0,2) and f (x) = 3x + 1, find f (11)

F(11)=F(3)=－F(－3)=－F(1)=－4

Increasing function, increasing function is increasing function, decreasing function is decreasing function, so what is the monotonicity of increasing function and decreasing function

If the increasing function grows fast and the decreasing function decreases slowly, it may be increasing
If the decreasing function decreases fast and the increasing function increases slowly, it may be decreasing
It may also be both increasing and decreasing, such as y = x ^ 2 + (- x)

The linear equation with a slope of - 1 / 2 and an intercept of 5 on the y-axis is?

y=k*x+b
k=-1/2,b=5
y=-1/2*x+5
x+2*y-5=0

Given the function f (x) = (2acos ^ 2x) + bsinxcosx, and f (0) = 2, f (π / 3) = 1 / 2 + radical 3
(1) If f (a) = 0, a belongs to (0,2 π), find the value of A

1、f(0)=2,f(π/3)=1/2+√3/2.
Substitute the function (x) = (2acos ^ 2x) + bsinxcosx to get a = 1, B = 2
f（x）=2cos²x+2sinxcosx
=sin2x+cos2x+1
=√2sin(2x+π/4)+1
The minimum value of the function = - radical 2, the maximum value = radical 2
2. F (a) = 0, a belongs to (0,2 π),
f（a）=√2sin(2a+π/4)+1=0
Sin (2a + π / 4) = - radical 2 / 2
2A + π / 4 = π - π / 4 or 2 π + π - π / 4
A = π / 4 or 5 π / 4

Urgent need for 35 inequalities and 35 binary linear equations

The line L passing through point P (2,3 / 2) intersects with the positive half axis of X and the positive half axis of Y respectively at a, B and o as the origin of coordinates. The area of triangle AOB is equal to 6 gods help
The line L passing through point P (2,3 / 2) intersects with the positive half axis of X and the positive half axis of Y respectively at a, B and o as the origin of coordinates. The area of triangle AOB is equal to 6. The equation of line L is obtained

Let l be Y-3 / 2 = K (X-2), K

The known function f (x) = a-2-a-2cos (2wx + 2 φ) (a is greater than 0, W is greater than 0, 0 ″φ ″π - 2)
And the maximum value of y = f (x) is 2, the distance between the two adjacent symmetrical axes of the image is 2, and pass through the point (1,2) 1. Calculate φ 2. Calculate f (1) + F (2) + F (2008)

The maximum value of y = f (x) is 2, so a / 2 + A / 2 = 2 leads to a = 2, and the distance between two adjacent symmetrical axes of the image is 2, so the minimum positive period T = 4, that is, 2 π / (2W) = 4 leads to w = π / 4, so f (x) = 1-cos (π / 2x + 2 ψ) brings in the point (1, 2) We get cos (π / 2 + 2 ψ) = - 1 and 0 ≤ ψ ≤ π / 2, and then we get ψ = π / 4, f (1) = 2F (2) = 1-cos (π + π / 2) = 1F (3) = 1-cos (3 π / 2 + π / 2) = 0f (4) = 1-cos (2 π + π / 2) = 1, so f (1) + F (2) + F (3) + F (4) = 4, so f (1) + F (2) + '` f (2008) = 502 f (1) + F (2) + F (3) + F (4)] = 2008

Famous sayings, aphorisms, sayings or poems about "cooperation"

The greatest strength is unity. (Uzbek) 002 is not afraid not to turn over, just afraid not to be united. 003 success lies in harmony, strength lies in unity. (Uygur) 004 arrows are full of bags, elephants step on constantly, the strength of unity is better than elephants. (DAI) 005 help each other when climbing mountains and crossing rivers

The congeners are: (a + b) 2 − 32 (a + b) − 14 (a + b) 2 + (− 2) 3 (a + b)

The original formula = (a + b) 2-32 (a + b) - 14 (a + b) 2-8 (a + b) = 34 (a + b) 2-192 (a + b)

Given that the two roots of square + C = 0 of quadratic function ax of one variable are - 2 and 3, and a > 0, find the solution set of square + BX + C > 0 of inequality ax

a> 0, ax & # 178; + BX + C opening upward
And ax & # 178; + BX + C > 0 is above the x-axis
Because x = - 2 and 3 are equal to 0
So make a sketch
Above the x-axis is X3