Find the range of sinx 0 x 2 of 5 4cosx under the root number of the function f x .

First of all, this is a continuous derivative function, and the range of values can be obtained by finding the boundary values as well as the extreme values.

Derivation. f'(x)

cosx root number (5+4cosx) +2(sinx) 2 (5+4cosx) (3 2).

cosx (5+4cosx) +2(sinx)^2]/(5+4cosx)^(3/2)

2(cosx)^2 + 5cosx + 2]/(5+4cosx)^(3/2)

2cosx + 1)(cosx + 2)/(5+4cosx)^(3/2)

So, when 2cosx + 1 = 0, i.e. cosx = -1 2.

f'(x) = 0

cosx = -1 2 correspondence.

x = 120 degrees sinx = 3 2

and x = 240 degrees sinx = - 3 2

At x = 120 degrees.

f(x) = (√3 /2)/√(5 - 4 *1/2) = 1/2

at x = 240 degrees.

f(x) = -1/2

And at the border.

f(0) = f(360 degrees) = 0

On [0<=x<=2 ], f(x) is a continuous function. Combined with the boundary value and the extreme value, it can be known.

f(x) increases monotonically at [0, 120], decreases monotonically at [120, 240], and increases monotonically at [240,360].

In summary, the range is .

sinx/cosx=tanx

When x belongs to (0 x 2).

tanx is a total real number.

Therefore, the number in the root number can be removed from all real numbers, because the number in the root number must be greater than or equal to 0fx, so the range is (0, positive infinity).

f(x) 2=(sinx) 2 (5+4cosx)=chain (cosx+let a=cosx+, so a belongs to [, so f(x) 2=if and only if x=x+9 16x has a minimum orange value, at which point f(x) 2 max= m....

This is a continuous derivative function, which can be obtained by finding the boundary value and the extreme value of the source of the mountain through the comma state, and the derivative f of the value range can be obtained'(x) =cosx root number(5+4cosx) +2(sinx) 2 (5+4cosx) (3 2)= cosx (5+4cosx) +2(sinx) 2] (5+4cosx) (3 2)= 2(cosx) 2 + 5cosx + 2] (5+4cosx) (3....)

f(x)=4sinx*2sin(x-π/3)-3=-4【cos(x+x-π/3)-cos(x-x+π/3)】-3

4cos(2x-π/3)-1

So when x [0, slag 2], 2x- 3 imitates the beam [Bishu-3,2 3].

So when x = 2, f(x) max = 1

When x=6, f(x)min=-5

So the range of f(x) is -5,1

This is a continuous derivative function, which can be obtained by finding the boundary value and the extreme value of the source of the mountain through the comma state, and the derivative f of the value range can be obtained'(x) =cosx root number(5+4cosx) +2(sinx) 2 (5+4cosx) (3 2)= cosx (5+4cosx) +2(sinx) 2] (5+4cosx) (3 2)= 2(cosx) 2 + 5cosx + 2] (5+4cosx) (3....)

f(x)=sinx-root number delay wheel 3cosx

2[sinx*(1 2)-cosx*( 3 2)]2[sinx*cos( 3)-cosx*sin( 3)]2sin(x- code envy 3).

x∈【-0】

x- 3 [-4 Paison 3, - 3].

sin(x-π/3)∈【1,√3/2】

So y [-2, 3].

The domain of the function f(x)=sinx (5+4cosx) (0 x 2).

Solution: Let f (x) = [cosx (5+4cosx)-sin(-4sinx) 2 (5+4cosx)] 5+4cosx).

cosx(5+4cosx)+2sin²x]/[5+4cosx)√(5+4cosx)]=0

Get cosx(5+4cosx)+2sin x=5cosx+4cos x+2sin x=2cos x+5cosx+2=(2cosx+1)(cosx+2)=0

Since cosx+2>0, there must be 2cosx+1=0, that is, cosx=-1 2, that is, the stationing point x = 3=2 3;

x₂=π3=4π/3；x is the maximum, and x is the minima.

f(x₁)=sin(2π/3)/√5+4cos(2π/3)]=sin(π/3)/√5-4cos(π/3)]=3/2)/√3=1/2

f(x₂)=sin(4π/3)/√5+4cos(4π/3)]=3/2)/√3=-1/2

Therefore, the value range is [-1 2,1 2].

First of all, this is a continuous derivative function rundown, and the range of values can be obtained by finding the boundary values and the extreme values.

Derivation. f'(x)

cosx root number (5+4cosx) +2(sinx) 2 (5+4cosx) (3 2).

cosx(5+4cosx)+2(sinx)^2]/(5+4cosx)^(3/2)

2(cosx)^2+5cosx+2]/(5+4cosx)^(3/2)

2cosx+1)(cosx+2)/(5+4cosx)^(3/2)

So when the god pants 2cosx+1=0, that is, cosx=-1 2.

f'(x)=0

cosx = -1 2 correspondence.

x = 120 degrees sinx = 3 2

And. x = 240 degrees sinx = - 3 2

At x = 120 degrees.

f(x)=(3/2)/√5-4*1/2)=1/2

at x = 240 degrees.

f(x)=-1/2

And at the border.

f(0) = f(360 degrees) = 0

On [0<=x<=2 ], f(x) is a continuous function. Combined with the boundary value and the extreme value, it can be known.

f(x) increases monotonically at [0, 120], decreases monotonically at [120, 240], and increases monotonically at [240,360].

In summary, the value range is hail intensity.