Answer:
Domain = -2, range = -5
Domain =>0, range = 0
domain = 2, range = 5
let y =>f(x)
It can be seen that Function can be linear having common difference 2
F(x) = 2x + 1 for x>0
is satisfied for above values
let's check
F(2) = 2(2)+1 = 4+1 = 5 As given
For x<0 , F(x) = 2x -1
f(-2)= 2(-2) -1= -4-1= -5 As given
For X= 0, F(x)= 2x
So , F(0) = 0 As given
