1. For x<3, h(x)=x+2
now x+2 is an increasing line (a practical way to check, h(1)=3, h(2)=4)
the largest value it takes is at 3, not inclusive, so h(3)=5 not inclusive.
Range_1= (-∞, 5)
2. for x> or equal to 3, h(x)=-x+8, which is a decreasing line (check h(4)=4, h(5)=3 )
so this line takes its maximal value at x=3, f(3)=5 and then takes any other value to -∞.
Range_2=(-∞,5]
3. Range(h)=Range1∪Range2=(-∞, 5)∪(-∞,5]=(-∞,5] (B)
