Given the function f (x) = cosx (SiNx cosx) + 1 (1), find the minimum positive period of F (x). (2) find the range of values. (3) find the monotone decreasing interval To complete the process

Given the function f (x) = cosx (SiNx cosx) + 1 (1), find the minimum positive period of F (x). (2) find the range of values. (3) find the monotone decreasing interval To complete the process

F (x) = (2cosxsinx-2cosxcosx + 1) / 2 + 1 / 2 = (sin2x cos2x) / 2 + 1 / 2 = √ 2 / 2 * sin (2x - π / 4) + 1 / 2T = π, maximum value √ 2 / 2 + 1 / 2, minimum value - √ 2 / 2 + 1 / 2

F (x) = the range and period of radical 3 * cos ^ 2x + sinxcosx Write the process in detail

f(x)=√3*cos^2x+sinxcosx
=√3*(1+cos2x)/2+sinxcosx
=√3/2(1+cos2x)+1/2sin2x
=sin(2x+∏/3)+√3/2
The value range is 〔 - 1 + √ 3 / 2,1 + √ 3 / 2], and the period is Π

Find the definition domain, range and monotone interval of the function y = log2 (- x ^ 2 + 2x + 15)

The function y = log2 (- x ^ 2 + 2x + 15) is meaningful
-x^2+2x+15>0
x^2-2x-15

Find the definition domain, range and monotone interval of function y = log2, (x-x 2)

If x ∈ - 0, X ∈ - 0, X ∈ - 1, then x ∈ - 0, X ∈ - 1 is defined,
Answer: the definition domain is interval (0,1)
Since the axis of symmetry of the parabola with the opening downward t = x-x? 2 is x = half, and the function T is an increasing function at x ∈ (0, half), and the function y = ㏒ 2 T is an increasing function,
Answer: the monotone interval is (0, half), and (half, 1). The interval function in the head is monotone increasing, and the interval at the end is decreasing

The monotone increasing interval of the function y = log2 (x2-6x + 5) is______ .

From x2-6x + 5 > 0, the solution is: x < 1 or x > 5,
U = x2-6x + 5 is monotonically decreasing on (- ∞, 1),
The required function is based on 2, according to the "same increase but different decrease",
Then the function y = log2 (x 2-6x + 5) increases the function on (5, + ∞)
The monotonic increasing interval of the function y = log2 (x2-6x + 5) is (5, + ∞)
So the answer is: (5, + ∞)

Given the function f (x) = log2 (x + 1) / (x-1) + log2 (x-1) + log2 (3-x), find the definition domain of function f (x); and the range of value

The definition domain is (1,3) f (x) = log2 [(x + 1) (3-x)] = log2 (- x 2 + 2x + 3) let t = - x 2 + 2x + 3. This is a quadratic function with the opening downward and the axis of symmetry x = 1. Because x belongs to (1,3), and t belongs to (0,4), then y = f (x) = log2 [(x + 1) (3-x)] = log2 (T), if t belongs to (0,4), then y belongs to (- ∞, 2)

F (x) is a function with t as its period. Find the periodic function of F (x) + F (2x) + F (3x) + F (4x),

F (2x) period is t / 2
F (3x) period is t / 3
F (4x) period is t / 4
So it is to find the least common multiple of T, t / 2, t / 3, t / 4
That is, the least common multiple of the molecule and the greatest common factor of the denominator
T is t / 1
So the least common multiple of the molecule is t
The greatest common factor of the denominator is 1
So the period of F (x) + F (2x) + F (3x) + F (4x) is t

What is the minimum period and maximum of the function f (x) = 3sin (2x + U / 4)

Wu 3

Given the function f (x) = 2sinscosx-2 radical sign 3sin ^ 2x, find the period, maximum and minimum value of F (x)

f(x)=2sinxcosx-2√3sin^2 x
f(x)=sin2x-√3(1-cos2x)
f(x)=sin2x-√3+√3cos2x
f(x)=2sin(2x+π/3)-√3
Period T = 2 π / |ω| = π
Maximum: 2 - √ 3
Minimum: - 2 - √ 3

Solution: find the minimum positive period and maximum value of the function f (x) = 5 √ 3cos ^ 2x + √ 3sin ^ 2x-4sinxcosx

f(x)=5√3(cos2x+1)/2+√3(1-cos2x)/2-2sin2x
=3√3+2√3cos2x-2sin2x
=3√3-4sin(2x-π/6)
The minimum positive period is π
The maximum value is 3 √ 3 + 4