Let even function f (x) satisfy f (x) = 2 & # 710; x-4 (x ≥ 0), then {x| f (X-2) > 0}= 1. {x | x 〈 - 2 or X 〉 4} 2. {x | x ＜ 0 or X ＞ 4} 3. {x | x ＜ 0 or X ＞ 6} 4. {x | x ＜ - 2 or X ＞ 2}

Let even function f (x) satisfy f (x) = 2 & # 710; x-4 (x ≥ 0), then {x| f (X-2) > 0}= 1. {x | x 〈 - 2 or X 〉 4} 2. {x | x ＜ 0 or X ＞ 4} 3. {x | x ＜ 0 or X ＞ 6} 4. {x | x ＜ - 2 or X ＞ 2}

x>=0
f(x-2)=2^(x-2)-4>0
2^(x-2)>4=2^2
x-2>2
x>4
x0
So f (- x) = 2 ^ (- x) - 4
Even function then f (x) = f (- x) = 2 ^ (- x) - 4
f(x-2)=2^[-(x-2)]-4>0
2^(-x+2)>2^2
-x+2>2
x

The even function f (x) decreases when x is greater than 0, f (1) = 0, and the value range of X is obtained when f (x) is less than or equal to 0

Even function, x > 0, decreasing then X

For the even function f (x), if x is greater than or equal to 0 and f (x) = X-1, then the solution set of F (x-1) less than 0 is?
Crowtaro, I want to ask, is the condition of "even function" still useful when using image method?

Because f (x) = X-1 (x > 0), so f (x-1) = X-2 (x > 0); because even function f (x), f (x) = X-1 (x > 0), so f (- x) = f (x) = X-1 (x < 0), so f (x-1) = X-2 (x > 0); f (x-1) = - x + 2 (x < 0); so we can find X-2 < 0 (x > 0)