A simple problem about function domain in senior one If the domain of y = f (x-1) is [2, + ∞), find the domain of y = f (x) I made it [3, + ∞), but I think I'm wrong, thanks

A simple problem about function domain in senior one If the domain of y = f (x-1) is [2, + ∞), find the domain of y = f (x) I made it [3, + ∞), but I think I'm wrong, thanks

What a mistake
Teach you a simple method, special value method
When x = 2, y = f (x-1) = f (1)
When x = 1, y = f (x) = f (1)
The definition domain is [1, + ∞)
This method is very easy to use, especially in the choice of filling in the blank, the speed is very fast
In fact, you can set y = f (x) = f (T)
This is more conventional

Range definition field
（1） If y = log2 (x * 2-2), and the value range of the function is [1, log2, 14], then the definition field of this function is ()
（2） The range of function f (x) = Log1 / 2 (x * 2-2x + 5) is ()
And the correct answer is [- 4, - 2] [2, 4] (- infinity, 27]

1.[2,4]

Some problems about the range of value and the domain of definition of higher one function,
The following three questions do not need steps, just tell the answer directly:
（1） The number of intersections between the image of the function y = f (x) and the line x = a is []
（2） The range of function y = SiNx / 2-sinx is []
（3） The range of (13-4x) under the function y = 2x-1-radical is []
）：
（1） Find the range of function y = (1 + SiNx) / (1 + cosx)
（2） Find the domain of the following functions:
(1) F (x) = [(x + 2) / (x + 4)]
(2) Y = [square of 25-x under root sign] + LG (cosx)
（3） Find the range of the power of (x + 2) - 3 * 4 of the function y = 2
I'm going to use it at 9:30, so the sooner, the better,
I don't know what the first question means

The first two questions on the second floor are wrong!
（1） 0 or 1
（2） [- 1-sqrt (2) / 2,1 + sqrt (2) / 2] Note: sqrt (2) / 2 is the root of 2
（3） Y