f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx) Finding the domain of F (x) and the monotone interval on (0,2)

f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx) Finding the domain of F (x) and the monotone interval on (0,2)

f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx)
=[(1+2sin2x)+(sinx+cosx)]/(1+sinx+cosx)
=[(sinx+cosx)^2+(sinx+cosx)]/(1+sinx+cosx)
Let t = SiNx + cosx = radical 2 * sin (x + 45), - radical 2

F (SiNx + cosx) = cosx + SiNx + sin2x-3 to find the maximum value of F (x)
I don't have any points. Please help me. Thank you

Let SiNx + cosx = t, t between positive and negative root sign 2!
(sinx+cosx)^2=1+2sinx*cosx=1+sin2x
So sin2x = (SiNx + cosx) ^ 2-1 = T ^ 2-1
So f (SiNx + cosx) = f (T) = t + T ^ 2-1-3 = T ^ 2 + T-4 = (T + 0.5) ^ 2-4.25
Then the condition that t is between the positive and negative root sign 2 is brought into the equation
The minimum value of T is - 0.5, the minimum value is - 4.25, and the maximum value of T is when the root sign is 2
I'm going back. I'll see the rest!

If f (SiNx) = sin3x, then f (cosx)=____

f(cosx)
=f[sin(π/2-x)]
=sin[3(π/2-x)]
=sin(3π/2-3x)
=-cos3x

1. F (SiNx) = sin3x for f (cosx)
2. F (x) is an odd function, and the domain of definition is R. when x is greater than 0, f (x) = x (1 + &# 179; √ x) to find f (x)

Two formulas sin (3x) = 3sinx-4sin ^ 3 (x), cos (3x) = 4cos ^ 3 (x) - 3cosx are given
These two formulas can be easily deduced
1, f (SiNx) = sin3x = 3sinx-4sin ^ 3 (x), then f (x) = 3x-4x ^ 3, domain [- 1,1]
Then, f (cosx) = 3cosx - 4cos ^ 3 (x) = - cos (3x)
2, when x = 0, f (0) = 0, when x0, substituting into the expression, f (- x) = - x (1 - & sup3; √ x), and by odd function, f (- x) = - f (x), so when x = 0, f (x) = x (1 + & sup3; √ x); when x

The range of function y = x + 1 / 4x on [1,2]

y=x+1/(4x)≥1
if and only if
X = 1 / (4x), that is, when x = 1 / 2
So the minimum is x = 1, y = 5 / 4
Maximum x = 2My = 17 / 8
So the range [5 / 4,17 / 8]

Find the range of function y = (x ^ 2 + 4x + 1) / (x ^ 2 + X + 1)

yx²+yx+y=x²+4x+1
(y-1)x²+(y-4)x+(y-1)=0
If x is a real number, the equation has a solution
So the discriminant is greater than or equal to 0
y²-8y+16-4y²+8y-4>=0
y²

Y = - x ^ 2-4x + 1, X ∈ [a, a + 1], find the range of function

f(x)=y=-(x+2)^2+5
The opening is downward and the axis of symmetry x = - 2
Endpoint value f (a) = - A ^ 2-4a + 1
f(a+1)=-a^2-6a-4
Vertex value f (- 2) = 5
If a

f(sinx)=2-2cosx,f(cosx)=?

From the title,
f(sinx)=2-2cosx,
therefore
f(cosx)
=f(sin(π/2-x))
=2-2cos(π/2-x)
=2-2sinx

If f (SiNx) = 3-cos2x, then f (cosx)=______ ．

F (cosx) = f [sin (π 2 − x)] = 3-cos (π - 2x) = 3 + cos2x

If f (SiNx) = 3-cos2x, then f (x)=

Let SiNx = t
cos2x=1-2sinx^2
=1-2T^2
f(sinx)=3-cos2x
f(T)=3-{1-2T^2}
So f (x) = 2x ^ 2 + 2