The function y = asin (Wx + ψ) + B (W 〉 0, |ψ||Π / 2) has the highest point (Π / 12,3) and the lowest point (7 Π / 12, - 5) in the same period Analytic formula of

The function y = asin (Wx + ψ) + B (W 〉 0, |ψ||Π / 2) has the highest point (Π / 12,3) and the lowest point (7 Π / 12, - 5) in the same period Analytic formula of

A=[3-(-5)]/2=4
b=[3+(-5)]/2= -1
Period T / 2 = 7 Π / 12 - Π / 12 = Π / 2, so t = Π, so w = 2 Π / T = 2
By substituting (Π / 12,3) into the function, we get: sin (Π / 6 + ψ) = 1, |Ψ|Π / 2, so Π / 6 + ψ = Π / 2, ψ = Π / 3,
Therefore, y = 4sin (2x + Π / 3) - 1

The function y = asin (Wx + φ) + B has the highest point (π / 11,3) and the lowest point (7 π / 12, - 5) in the same period

The period T = (7 π / 12 - π / 12) * 2 = π t = 2 π / w = π w = 2 a = (3 - (- 5)) / 2 = 4B = (- 5 + 3) / 2 = - 1 2 * π / 12 + φ = π / 2 + 2K π φ = π / 3 + 2K π φ = π / 3 + 2K π φ should have a range, for example: |||||||||||||||||||||||||
For adoption

The known function y = asin (ω x + φ) + C (a > 0, ω > 0, | φ| π) 2) In the same period, the coordinates of the highest point are (2,2), and the coordinates of the lowest point are (8, - 4) (1) Find the values of a, C, ω, and φ; (2) Find the monotone increasing interval of this function

（1）∵
A+C＝2
−A+C＝−4 ，∴
A＝3
C＝−1 ，
∵T=2（8-2）=12，∴ω=π
Six
∵3sin（π
6×2+φ）=3，∴π
6×2+φ=π
Two
∴φ=π
6．
   （2）∵-π
2+2kπ≤π
6x+π
6≤π
2+2kπ
∴-4+12k≤x≤2+12k
The monotone increasing interval of this function [- 4 + 12K, 2 + 12K] (K ∈ z)

It is known that y = asin (Wx + φ) + B (a > 0, w > 0, | φ| Pai / 2) is the highest in the same period The point is (2,2), and the lowest point is (8, - 4)

y=Asin(wx+φ)+B
A>0,w>0,|φ|

The known function y = asin (Wx + φ) (x ∈ R, 0 The second problem: find the monotone increasing interval of G (x) = f (x - π / 12) - f (x + π / 12)

(1) T = 2 (11 π / 12-5 π / 12) = π, then w = 2
F (5 π / 2) = 0 and 0

The function f (x) = asin (Wx + φ) (a > 0, w > 0, | φ) is known|

A:
1）
The lowest point Q (- π / 6, - 2), then a = 2
f(-π/6)=2sin(-wπ/6+b)=-2,sin(-wπ/6+b)=-1,b-wπ/6=-π/2
f(π/12)=2sin(wπ/12+b)=0,wπ/12+b=0
According to the above two formulas, w = 2, B = - π / 6
f(x)=2sin(2x-π/6)
2)
f(a+π/12)=2sin[2(a+π/12)-π/6]=3/8
sin2a=3/16,1+sin2a=19/16
(sina+cosa)^2=19/16
A is the third quadrant angle, Sina

How to do group a exercise 4 on page 44?

The solution set a gets a = {X / x = 1 or - 1}, because B is a subset of a, so B may be {X / - 1A = 1}, or {X / a = 1}, or an empty set, and a = 1 or - 1 or 0 is obtained

Answers to questions 1 and 2 of group A on page 39

1. (1) when (2.5, positive infinity), monotonically increasing
When (negative infinity, 2.5), it decreases monotonically
(2) When [0, positive infinity), monotonically decreasing
When (negative infinite, 0), monotonically increasing
2. (1) f (x1) - f (x2) = (x1 + x2) * (x1-x2)
(2) F (x1) - f (x2) = (x1-x2) / (x1 * x2)

Senior one mathematics compulsory 1 Chapter 2 exercises 2. PEP, quick to find ① (2) and (4) of pep

It's not over ~ only 1,2
One
（1）100
（2）-0.1
（3）4-π
（4）x-y
Two
（1）b3|2 a1|2
—— * —— =1
a1|2 b3|2
（2）√a
（3）m1|2*m1|3*m1|4
————————=1
m5|6*m1|4

Senior high school mathematics compulsory 1 Chapter 1 exercises 1

Group a 1. (1) {x | x ≠ 4} (2) x ∈ R (3) {x | x ≠ 1 and X ≠ 2} (4) {x | x ≤ 4 and X ≠ 1} 2. (1) not equal because of different definition domains (2) unequal because of different definition domains (3) equal 3. (1) definition domain R (2) definition domain {x ≠ 0} value domain {y ≠ 0} (3) definition domain R