The number of zeros of F (x) = lgx + X + 1

The number of zeros of F (x) = lgx + X + 1

The number of zeros is f (x) = 0
lgx+x+1=0
lgx=-x-1
There is only one zero point in the observation of combination of number and shape

The minimum positive period of y = SiNx (2sinx + 2cosx) is

y=sinx(2sinx+2cosx)
y=2sin^2x+2sinxcosx
y=2sin^2x+sin2x
=1-cos2x+sin2x
=√2sin（2x-π/4）+1
So the minimum positive period of y = SiNx (2sinx + 2cosx) is t = 2 π / 2 = π

Find the minimum positive period of (1) y = √ 3 / 2sinx-1 / 2cosx; (2) y = SiNx + 2cosx

（1）y=√3/2sinx-1/2cosx = sinx cos∏/6 - cosx sin∏/6 = sin(x - ∏/6)；
Therefore, Tmin + = 2 Π / w = 2 Π / 1 = 2 Π
(2) Y = SiNx + 2cosx = √ 5 sin (x + α), where Tan α = 2
Tmin+ = 2∏/w = 2∏/1 = 2∏

Y = 2sinx ^ 2 minimum positive period

Y = 2Sin & sup2; X = 1-cos2x  minimum positive period T = π (cos2x = 1-2sin & sup2; X, t = 2 π / ω of cos ω x)

What is the minimum positive period of y = 1 / 2sinx - 2
What is the minimum positive period of y = half sin square x

y=0.5sin²x
=0.5*(1-cos2x)/2
=0.25(1-cos2x)
Know,
T=2π/2=π,
The minimum positive period is π

Given that f (x) = 2sinx / 2cosx / 2 + 2cosx / 2-1, find the set of all real numbers x that make f (x) + F '(x) = 0, urgent!
It should be f (x) = 2sinx / 2cosx / 2 + 2 (cosx / 2) ^ 2-1

f(x)=2sin(x/2)cos(x/2)+2cos²(x/2)-1
=sinx+cosx
f'(x)=cosx-sinx
f(x)+f'(x)=2cosx
Because f (x) + F '(x) = 2cosx = 0
So x = k π + π / 2 (K ∈ z)

Given f (x) = 2sinx & # 47; 2cosx & # 47; 2 + 2cos ^ X & # 47; 2-1, find all sets that make f (x) + F & # 39; (x) = 0 hold

Simplification: F (x) = SiNx + cosx
f'(x)=cosx-sinx
From F (x) + F '(x) = 2cosx = 0
We get: x = k π + π / 2, K is any integer
The set is {x | x = k π + π / 2, K ∈ Z}

Given the function f (x) = 1 / 2cosx ^ 2 + √ 3 / 2sinxcosx + 1, X belongs to R
If √ is the root sign, then √ 3 is divided by 2
1. When the function y reaches the maximum value, the set of independent variables X is obtained; 2. The image of the function can be obtained by how the image of y = SiNx (the domain of definition is R) is stretched and shifted

1.1/2cosx^2+√3/2sinxcosx+1=-1/4cos2x+√3/4sin2x+5/4=
1/2sin(2x-30)+4/5
2x-30=360N+90 max
x=60+180N
2. The function image is reduced by 1 / 2 from y = SiNx, extended by 2 times from X axis, and shifted 30 degrees to the right of X axis/

If the sum of the maximum and minimum values of the function y = ax (a > 0 and a ≠ 1) in the interval [0, 2] is 3, then a=______ ．

The monotonicity of ∵ function y = ax (a > 0 and a ≠ 1) is consistent. When a > 1, the function f (x) increases monotonically in the interval [0,2]. When 0 < a < 1, the function f (x) decreases monotonically in the interval [0,2], and the sum of the maximum and minimum of the function is 3, ∵ f (2) + F (0) = A2 + 1 = 3. The solution is a = - 2 (rounding off), or a = 2

If the function y = x2-2x + 3 has a maximum value of 3 and a minimum value of 2 in the interval [0, M], then the value range of M is ()
A. [1，∞）B. [0，2]C. （-∞，2]D. [1，2]

It can be seen from the meaning that the axis of symmetry of the parabola is x = 1, with the opening upward ∵ 0 on the left side of the axis of symmetry ∵ the image on the left side of the axis of symmetry is monotonic decreasing ∵ when x = 0 on the left side of the axis of symmetry, there is a maximum value of 3 ∵ and a minimum value of 2 on [0, M]. When x = 1, the value range of y = 2 ∵ m must be greater than or equal to 1 ∵ the image of the parabola about x = 1 symmetry ∵ m must be ≤ 2, so D is selected