Let y = 3 + sin x, y ∈ [2, 4] and assumes all value there in.

Also, let g(y) = y + , this function is increasing on [2, 4]

So, g(2) ≤ g(y) ≤ g(4)

Thus 3 ≤ g(y) ≤ and both extreme values are attained.

It now follows that the minimum of

f(b) = max ( |g(y) + b – 3|) is 3/4, which is attained at

b = – 3/4 for if b > – ,

then choose x = ,

So y = 4 and then g(y) + b – 3 > 3/4

While if b < , then chose x = – ,

So y = 2 and g(y) + b – 3 = – 3/4

On the other hand, range for g(y) is

≤ g(y) + b – 3 ≤ for b = –3/4