If the independent variable x takes two different values X1 and X2, and the function values are equal, then X1 + X2 is equal

If the independent variable x takes two different values X1 and X2, and the function values are equal, then X1 + X2 is equal

y=2x2-6x+3
Axis of symmetry = 6 / 4 = 3 / 2
therefore
x1+x2=3/2

Given the functions f (x) = 2lx ml and G (x) = XLX ml + 2m-8, if x 1 belongs to (negative infinity, 4), there exists x 2 belonging to [4, positive infinity], such that f (x) = XLX ml + 2m-8
Let f (x1) = g (x2) hold, and find the value range of real number m?

In order to make 2lx1 ml = x2lx2 ml + 2m-8, x1 ∈ (- ∞, 4), X2 ∈ [4, + ∞) when m > 4, x1-m4, when x2 = m, x2lx2 ml = 0, so x2lx2 ml + 2m-8 ∈ [2m-8, + ∞), f (x1) and G (x2) have the same range, so when m > 4, f (x1) = g (x2) can hold, when M0, so g (x2) has two zeros, so g (x2)

Given that the function f (x) = asin (2wx + θ) (W > 0) has the maximum value 2 when x = π / 12, X1 and X2 are any two elements in the set M = {x ∈ R | f (x) = 0}, and the minimum value of | x1-x2 | is π / 2
(1) Find f (x) (2) if f (α) = 2 / 3, find the value of COS (7 π / 6-2 α)

(1) f(x)=Asin(2ωx+θ) (ω>0)A=2T=π/2*2=π2ω=2π/T=2π/π=2ω=1f(x)=2sin(2x+θ)2=2sin(2*π/12+θ)2*π/12+θ=π/2θ=π/3f(x)=2sin(2x+π/3)(2) f(α)=2/32sin(2α+π/3)=2/3sin(2α+π/3)=1/3sin2αcosπ/3...

The function f (x) = - 3x ^ 2 + 2x-5 / 3 asks if there is f (x1) * f (x2) = - 1, if there is, then find X1 x2
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Does not exist in the range of real numbers
Because its function value is less than zero, it's Delta