Y = log2 (X & # 178; - 2x + 3) On the function FX = 4sin (2x + π / 3), (x ∈ R) Minimum positive period, symmetry point, symmetry axis

Y = log2 (X & # 178; - 2x + 3) On the function FX = 4sin (2x + π / 3), (x ∈ R) Minimum positive period, symmetry point, symmetry axis

(1) X & # 178; - 2x + 3 > X & # 178; - 2x + 1 + 2 > 0 (x-1) &# 178; + 2 > 0, the definition field is R (2) t = x & # 178; - 2x + 3 = x & # 178; - 2x + 1 + 2 = (x-1) &# 178; + 2 when x = 1, t gets the minimum value of 2; when x = 1, t has no maximum value, then y = Log & # 8322; (X & # 178; - 2x + 3) ≥ Log & # 8322; 2 = 1, the value field is [1, +

When 1

y=log2 [-(x-1)^2+4]
one

Inverse function of y = log2 ^ 2x

y=log2^2x
=>2^y=2x
=>x=2^y/2
=>x=2^(y-1)
So the inverse function is Y2 ^ (x-1) (x ∈ R)

Inverse function of y = log2 (1-2x)

2^y=1-2x

What is the inverse of the function y = - 2x + 5?

y=-2x+5
2x=5-y
x=(5-y)/2
So the inverse function: y = (5-x) / 2

Inverse function of function y (x) = 2x-1 / 2x + 1
Please be more specific. Thank you

I use MATLAB software to solve the problem as follows:
solve('y=2*x-1/(2*x)+1','x')
ans =
y/4 + (y^2 - 2*y + 5)^(1/2)/4 - 1/4
y/4 - (y^2 - 2*y + 5)^(1/2)/4 - 1/4

Find the inverse function of y = x Λ 2-2x + 3 (x ＜ = O),

y=x∧2-2x+3=(x-1)^2+2
X = 1-radical (Y-2)
The inverse function of y = x Λ 2-2x + 3 (x ＜ = O) is
Y = 1-radical (X-2) (x > = 2)

Y=-x^2+2X+3 X

Y = (X-2 under radical) - 1 6 = > x > = 2

The inverse function of y = 2 ^ (x ^ 2-2x + 3), X ∈ (1, + ∞)
Ask the master to solve the problem

If y = 2 ^ (x ^ 2-2x + 3) = 2 ^ [(x-1) ^ 2 + 2], X ∈ (1, + ∞), then Y > 2 ^ 2 = 4, log (2) (y) = (x-1) ^ 2 + 2log (2) (y) - 2 = (x-1) ^ 2, because x > 1, so √ [log (2) (y) - 2] = x-1x = 1 + √ [log (2) (y) - 2], so its inverse function is y = 1 + √ [log (2) (x) - 2], X ∈ (4, + ∞)

Finding the inverse function of y = x ^ 2-4x + 2 (x ≥ 2)

y=(x-2)²-2
(x-2)²=y+2
x>=2
So X-2 = √ (y + 2)
x=2+√(y+2)
The inverse function is y = 2 + √ (x + 2), X ≥ - 2