Let f (x) = asin (ω x + φ), X ∈ R (where a > 0, ω > 0,0)

Let f (x) = asin (ω x + φ), X ∈ R (where a > 0, ω > 0,0)

A=2
π=2π/ω,
ω=2
2(2π/3)+φ=3π/2
φ=π/6
f(x)=2sin(2x+π/6),
The period is π
And T = 2 π / ω = π
ω=2
∴f(x)=Asin(2x+φ)
The lowest point is m (2 π / 3, - 2)
∴A=2
sin(2*2π/3+φ)=sin(4π/3+φ)=-1
When x = 2 π / 3, there is a minimum
That is, 2 * 2 π / 3 + φ = 2K π - π / 24 π / 3 + φ = 2K π - π / 2
φ = 2K π - π / 2-4 π / 3 = 2K π - 11 π / 6 = 2... Expansion
The period is π
And T = 2 π / ω = π
ω=2
∴f(x)=Asin(2x+φ)
The lowest point is m (2 π / 3, - 2)
∴A=2
sin(2*2π/3+φ)=sin(4π/3+φ)=-1
When x = 2 π / 3, there is a minimum
That is, 2 * 2 π / 3 + φ = 2K π - π / 24 π / 3 + φ = 2K π - π / 2
φ=2kπ-π/2-4π/3=2kπ-11π/6=2kπ-2π+π/6=2(k-1)π+π/6
∵0＜φ＜π/2
∴φ=π/6
Ψ f (x) = 2Sin (2x + π / 6) retract
Given a function f {x} = asin [omiga x + Fai], then an image is given, and the minimum positive period, the analytic expression of this function and the monotone interval are obtained?
What a magic question
2 PA / Omega after the period T is known in the image = t to calculate Omega
In addition, the distance from the highest or lowest point of the image to the balance position is a
Then the abscissa of the first oblique upward focus is denoted as X, then omiga x + Fai = 0. By adding and subtracting a number of periods, we can get Fai in accordance with the range. The analytic expression of this function is given
-Handkerchief / 2
Let f (x) = piecewise function {① 2 (1-x) power minus a (x is less than or equal to 0); ② f (x-1), x > 0.}, if f (x) = x, there are only two
Let f (x) = piecewise function {1-x) minus a (x is less than or equal to 0);
If f (x) = x, and there are only two real roots, then the value of real number a is?
Drawing problem
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The function f (x) is a periodic function with a minimum positive period of 1 in the X-1 part
We can see that a = 2 is the critical point, and the function is the same as X
If the image of the positive scale function is parallel to the straight line y = - 23x + 4, the analytic expression of the positive scale function is___ .
Let y = KX (K ≠ 0), ∵ be parallel to the straight line y = - 23x + 4, ∵ k = - 23, ∵ be y = - 23x
Let f (x) = piecewise function {1-x power of 2 minus a (x is less than or equal to 0); ② f (x-1), x > 0.}, if f (x) = x, there are only two functions
Then you have 0 in your answer
My negligence
I drew a sketch to solve it
In a
The image of the positive proportion function is parallel to the straight line y = 3 / 3 - 2x + 4. What are the analytic expressions of the positive proportion function
There are many, as long as k = - 2 / 3
For example: y = - 2 / 3x + 8
Y = - 2 / 3x-12 and so on
As long as the coefficient of X is negative by two-thirds, it is OK?
The function f (x) = logarithm of base 4 (x power of 4 + 1) - X / 2 is known
If the function has and only has one common point with the logarithm of G (x) = base 4 (the x-th power of a2-4a / 3), find the value range of real number a
Don't believe Baidu's answer. Let f (x) - G (x) = log4 [(4 ^ x + 1) / a (2 ^ x-4 / 3)] - 1 / 2x = 0 (a ≠ 0), that is, (4 ^ x + 1) / a (2 ^ x-4 / 3) = 2 ^ x, let 2 ^ x = t (T > 0) let H (T) = 4 ^ x + 1-A * 2 ^ x (2 ^ x-4 / 3) = (1-A) T ^ 2 + 4 / 3At + 1 (T > 0, a (T-4 / 3) > 0, a ≠ 0) 1 'if a = 1, then H (t
Given that the ratio coefficient of a positive proportion function is - 5, its analytical expression is ()
y=-5x
y=-5x
lg25+2lg2+lg5*lg20+lg^22=?
The original formula = lg5 & # 178; + 2lg2 + lg5 × (lg5 + LG4) + LG & # 178; 2
=2lg5+2lg2+lg²5+lg5×lg4+lg²2
=2×(lg5+lg2)+lg²5+lg5×lg2²+lg²2
=2×lg10+lg²5+2lg5×lg2+lg²2
=2+(lg5+lg2)²
=2+lg²10
=2+1
=3
It is known that the positive proportion function passes through the point (- 1,2), and the proportion coefficient of this positive proportion function is_____
-2