Let f (x) = – 2x + 2, F 1 (x) = f (x), FN (x) = f [fn-1 (x)], n ≥ 2, n ∈ n, then the image of function y = FN (x) is always over the fixed point

Let f (x) = – 2x + 2, F 1 (x) = f (x), FN (x) = f [fn-1 (x)], n ≥ 2, n ∈ n, then the image of function y = FN (x) is always over the fixed point

Solve the fixed point equation: F (x) = - 2x + 2 = x to get: x = 2 / 3
So the function is always over the fixed point (2 / 3,2 / 3)

f（x）=|2x-1|，f1（x）=f（x），f2（x）=f（f1（x）），… , FN (x) = f (fn-1 (x)), then the number of zeros of the function y = F4 (x) is___ ．

We can get y = F4 (x) = f (F3 (x)) = | 2f3 (x) - 1 |, let it be 0, we can get F3 (x) = 12, that is, f (F2 (x)) = | 2F2 (x) - 1 | = 12, we can get F2 (x) = 34 or F2 (x) = 14, that is, f (F1 (x)) = 34 or 14, and f (F1 (x)) = | 2F1 (x) - 1 |, we can make it equal to 34 or 14

Given the function f (x), X ∈ R, and F 1 (x) = 2, f (f (f (x ））=FN (x) [where n is several, there are several F], find F4 (x) =?

Because x ∈ R, no matter which real number n takes, the result is 2

Given the sequence an = 1 / {(n + 1) square} FN = (1-a1) * (1-a2) *. * (1-an) by calculating the value of F1, F2, F3, infer the value of FN needs to be proved

f1=3/4
f2=2/3
f3=5/8
fn=(1-a1)*(1-a2)*.*(1-an)
=(1-1/2^2)(1-1/3^2)(1-1/4^2)*.*[1-1/(n+1)^2]
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)*.*[1-1/(n+1)][1+1/(n+1)
=1/2*3/2*2/3*4/3*3/4*5/4*.*n/(n+1)*(n+2)/(n+1)
=1/2*(n+2)/(n+1)
=(n+2)/(2n+2)

Given f (x) = x / 1-x, let F1 (x) = f (x), FN (x) = fn-1 [fn-1 (x)] (n > 1 and N ∈⊥ positive integer), find the expression of FN (x) (n ∈ positive integer)
The answer can be written on the paper in the form of an answer question (not the kind of solving F1, F2, F3, F4, and then guessing). That is to say, FN (x) can be directly deduced by some method,

f1(x)=f(x)=x/1-x=x/(1-x)f2(x)=f1(f1(x))=f1(f(x)=f1(x/1-x)=(x/1-x)/[1-(x/1-x)]=x/(1-2x)f3(x)=f2(f2(x))=f2(x/1-2x)=(x/1-2x)/[1-2(x/1-2x)]=x/(1-4x)f4(x)=f3(f3(x))=f3(x/1-4x)=(x/1-4x)/[1-4(x/1-4x)]=x/(1-8...

The value range of function (x) = √ 3sinxcosx sin ^ 2x-3 / 2 on [- π / 2,0] is -?

f(x)=√3/2 * sin2x+cos2x/2-2
=sin(2x+π/6)-2
x∈[-π/2,0]
2x+π/6∈[-5π/6,π/6]
sin(2x+π/6)∈[-1,1/2]
f(x)∈[-3,-3/2]

Find y = sin square x + 2cosx + 1, X belongs to [0, π / 2] range

Y = sin square x + 2cosx + 1
=1-cos²x+2cosx+1
=-(cos²x-2cosx+1)+3
=-(cosx-1)²+3
X belongs to [0, π / 2]
So 0 ≤ cosx ≤ 1
So the range is 2 ≤ y ≤ 3

The range of the function y = cos2x SiNx is______ ．

Function y = cos2x SiNx = 1-sin2x-sinx = - (SiNx + 12) 2 + 54, so when SiNx = - 12, function y has the maximum value of 54, when SiNx = 1, function y has the minimum value of - 1. So the range of function y is [− 1, 54], so the answer is: [− 1, 54]

How to find the range of y = (COS x + 1) / (COS X - 1)

Using double angle formula
cos x +1=2(cosx/2)^2
cos x -1=-2(sinx/2)^2
The original formula = - [(cosx / 2) / (SiNx / 2)] ^ 2 = - (Cotx / 2) ^ 2
Because the Cotx / 2 range is from negative infinity to positive infinity
So the range is y

F (x) = - cos ^ 2x + simx + 2, the range of F (x) is

F (x) = - [1-sin & sup2; x] + SiNx + 2 = Sin & sup2; X + SiNx + 1 = [SiNx + (1 / 2)] & sup2; + 3 / 4. Since - 1 ≤ SiNx ≤ 1, then - 1 / 2 ≤ SiNx + (1 / 2) ≤ 3 / 2, so 0 ≤ [SiNx + (1 / 4)] & sup2; ≤ 9 / 4, then the range of the original function is [3 / 4,3]