Let f (x) = vector a · vector B-1, where vector a = (2cosx, 1), vector b = (cosx, √ 3, sin2x), X ∈ R Finding the decreasing interval of F (x)

Let f (x) = vector a · vector B-1, where vector a = (2cosx, 1), vector b = (cosx, √ 3, sin2x), X ∈ R Finding the decreasing interval of F (x)

F (x) = 2 (cosx) ^ 2 + root 3sin2x - 1
=Root 3sin2x + cos2x
= 2sin(2x + π/6)
Let 2K π + π / 2 < 2x + π / 6 < 2K π + 3 π / 2
The decreasing interval of the solution is
（kπ+ π/6 ,kπ + 2π/3）

Given that the domain of F (x2-2) is [- 2,3], find the domain of F (x)

The domain of a function is the set of values of the independent variable X that makes the function meaningful
The domain of F (X & sup2; - 2) is [- 2,3],
х X & sup2; - 2, - 2 ≤ x ≤ 3,
0≤x²≤9,
-2≤x²-2≤7,
The domain of F (x) is [- 2,7]

Given that the domain of F = (x2-1) is [- 1,3], find the domain of F (x)

In the condition of y = f (T) t = x ^ 2-1, the domain of F = (x2-1) is [- 1,3], which refers to the range of X, that is: - 1 ≤ x ≤ 3 = = > 0 ≤ x ^ 2 ≤ 9 = = > - 1 ≤ x ^ 2-1 ≤ 8, that is, t ∈ [- 1,8]. So the domain of F (T) is [- 1,8]

It is known that the domain of function f (x 2-2) is the domain of [1,3] finding function f (x)
Wrong input, the domain of F (x + 2)!

The domain of definition in the question is the value of X in X & sup2; - 2, and the domain of X & sup2; - 2 in the domain [1,3] is the domain of F (x)

If the domain of F (x + 1) is [- 1 / 2,2], find the domain of F (x2)
But that's the answer
∵－1/2＜x＜2
∴1/2＜x＋1＜3
∴1/2＜x2≤3
So x is greater than or equal to the negative root sign three and less than the negative half root sign two
Or X is greater than half of the root sign 2 and less than or equal to the root sign 3
But why is x + 1 x2? I don't understand

The domain of F (x + 1) is [- 1 / 2,2]
That is - 1 / 2=

If the function f (x) is continuous on [0,1] and f (0) = f (1), then for any natural number n, there exists ξ ∈ [0,1], such that f (ξ + 1 / N) = f (ξ)!

Let f (x) = f (x + 1 / N) - f (x)
F(0)=f(1/n)-f(0)
F(1/n)=f(2/n)-f(1/n)
…
F[(n-1)/n]=f(1)-f[(n-1)/n]
Then f (0) + F (1 / N) + +F[(n-1)/n]
=f(1/n)-f(0)+f(2/n)-f(1/n)+… +f(1)-f[(n-1)/n]
=f(1)-f(0)
=0
So f (0) = f (1 / N) = F [(n-1) / N] = 0 or the sign of F (I / N) and f (J / N) is opposite (0 ≤ I)

F (n) = 1 + 1 / 2 + 1 / 3 +... 1 / N, is there a function g (n) about natural number n, so that the equation f (1) + F (2) +... + F (n-1) = g (n) × [f (n) - 1] holds for all natural numbers n ≥ 2? And prove your conclusion

Given that the domain of F (x) is a set of natural numbers n, and satisfies the condition f (x + 1) = f (x) + F (y) + XY, and f (x) = 1, find f (x)

Let y = 1, then f (x + 1) - f (x) = f (1) + x = 1-x
When x = 1, f (2) - f (1) = 2
When x = 2, f (3) - f (2) = 3
…… ……
f(x)-f(x-1)=x ＠
Put ① + ② + +@There are:
f(x)-f(1)=2+3+…… x
f(x)=x(x+1)/2
This is not my school homework this week

What is the range of the function y = (2-x) / (x + 1)?

This problem can be solved by using the inverse representation method or the method of calculating the range by separating constants,
1、 Y = 2-x / x + 1
y(x+1)=2-x
yx+y=2-x
x(y+1)=2-y
x=2-y/y+1
Because y + 1 is not equal to 0
So y is not equal to - 1

Given the function f (x) = 1 / 2 of X + X, judge the monotonicity of F (x) on (0,1] and prove it

Monotonic decreasing of F (x) on (0,1]
Set 0