Let f (x) = 2x + SiNx radical 3cosx, the tangent slope of the image of function f (x) at M (x0, fX0)) is 2

Let f (x) = 2x + SiNx radical 3cosx, the tangent slope of the image of function f (x) at M (x0, fX0)) is 2

Because f (x) = 2x + SiNx - √ 3cosx
So f '(x) = 2 + cosx + √ 3sinx = 2
So cosx + √ 3sinx = 0
2 sin (x + π / 6) = 0
So sin (x + π / 6) = 0
So x + π / 6 = 2K π (K ∈ z)
So x = 2K π - π / 6 (K ∈ z)

The function f (x) = the slope of tangent at point (1, √ 2) under root sign (1 + x)

Seeking guidance
f(x)=(x^2+1)^(1/2)
f'(x)=(1/2)*(x^2+1)^(1/2-1)*(x^2+1)'
=(1/2)*(x^2+1)^(-1/2)*2x
=x/√(x^2+1)
X=1
f'(x)=1/√2
So the tangent slope is equal to √ 2 / 2
y-√2=√2/2*(x-1)
x-√2*y+1=0

Function 5 * root sign ((x Λ 2) + 400) calculates the abscissa of tangent point with slope 3

Fifteen

Given the function f (x) = asin (x + φ) (a > 0, 0 < φ < π), the maximum value of X ∈ R is 1, and its image passes through the point m (π) 3，1 2)． (1) Find the analytic expression of F (x); (2) α, β∈ (0, π) is known 2) And f (α) = 3 5，f(β)＝12 13, find the value of F (α - β)

(1) If a = 1, then f (x) = sin (x + φ). By substituting the point m (π 3, 12), we get sin (π 3 + φ) = 12, and 0 ＜ φ ＜ π, νπ 3 + φ = 56 π, νφ = π 2, so f (x) = sin (x + π 2) = cosx. (2) cos α = 35, cos β = 1213, and α, β∈ (0, π 2)

The function y = asin (Wx + φ) (where a > 0, w > 0, the absolute value of φ) is known

T=π=2π/w--> w=2
The ordinate of the highest point is 3 / 2 --- a = 3 / 2
The equation of symmetry axis is x = π / 6 >, because the axis of symmetry of sin function is on π / 2 + K π, so φ = - π / 6 + K π + π / 2 --- > φ = π / 3
y=1.5sin(2x+π/3)
The increasing range is 2K π - π / 2=

The maximum value of the function f (x) = asin (Wx - π / 6) + 1 (a > 0, w > 0) is 3, and the distance between two adjacent symmetry axes of the image is a half of the derivative The maximum value of the function f (x) = asin (Wx - π / 6) + 1 (a > 0, w > 0) is 3, and the distance between the two adjacent symmetry axes of the image is derived by half. (1) find the analytic formula of the function f (x) (2) let a belong to the school of zero to half, then f (A / 2) = 2, calculate the value of A

If a = 3-1 = 2, t = Pai / 2 * 2 = Pai, then w = 2pai / T = 2
Therefore, f (x) = 2Sin (2x Pai / 6) + 1
f(a/2)=2sin(a-Pai/6)+1=2
sin(a-Pai/6)=1/2
a-Pai/6=Pai/6
a=Pai/3

Let f (x) = asin (Wx + FAI) (where x belongs to R, a > 0, w > 0, - Pai / 2)

Verification: Verification: Verification: when x = 1, y = 1, y = 1, y = 1, x = 1, x = 1, x = 1, a = 1, a = 1, a = 1, a = 1, a = 1, a = a = 1, a = a = 1, a = a = 1, a = 1, a = a = 1, a = 1, a = 1, a = 1, a = 1, a = 1, a = 1, a = 1, a, a, a, a, a, a, a, a, a, a, a, a, a, a = a = 1, a = 1, f (2) = 1, f (x = x) = Sinn (π / 4) x + (π / 4) x + (π / 4, 4, 4, 4, verify: Verification: Verification: x = 1, y = through the above calculation, we know that f (- 1) = 0, f (1) = 1, f (...)

Given the function y = asin (ω x + φ), in the same period, when x = π When x = 7 π If y = - 2, then the analytic formula of the function is () A. y=1 2sin（x+π 3） B. y=2sin（2x+π 3） C. y=2sin（x 2-π 6） D. y=2sin（2x+π 6）

The function y = asin (ω x + φ), in the same period, when x = π
When x = 7 π
At 12, the minimum value y = - 2 is obtained,
So a = 2,
ωπ
12+Φ=π
2，ω7π
12+Φ=3π
Two
The solution is: ω = 2
φ=π
Three
The analytic formula of the function is y = 2Sin (2x + π)
3）
Therefore, B is selected

Let y = asin (Wx + Fei) + B (a > 0, w > 0, | Fei)|

It is known that in the same period, when x = 5 π / 3, y has a maximum value of 7 / 3; when x = 11 π / 3, y has a minimum value of - 2 / 3, then t / 2 = 11 π / 3-5 π / 3 = 2 π, then the period T = 2 π / w = 4 π can be obtained. Therefore, the solution w = 1 / 2 and the function has the maximum value a + B = 7 / 3 when sin (Wx + φ) = 1

The known function FX = asin (Wx + φ) (x ∈ R, a > 0, w > 0,0) is known

A=2
T=4*[π/6-(-π/6)]=4π/3w=2π/(4π/3)=1.5f(x)=2sin(1.5x+φ)2sin(1.5*π/6+φ)=2π/6+φ=π/2φ=π/3
f(x)=2sin(1.5x+π/3)
g(x)=[f(x-π/12)]^2
=[2sin(1.5(x-π/12)+π/3)]^2
=[2sin(1.5x+π/12)]^2
=2-2cos(3x+π/6)
-π/6≤x≤π/3
-π/3≤3x+π/6≤7π/6
3x+π/6=π
x=5π/18
ymax=4