Given,
Y = log[sin(x-3)]+ √(16-x^2) = f(x)+ g(x);
Let, f(x)= log[sin(x-3)] & g(x)= √(16-x^2); Domains of log() funcn & √() funcn are different.
To find Y domain find f(x), g(x) domains it's intersection interval gives the answer i.e; Domain of Y.
Finding domain of f(x) =log[sin(x-3)]
We know that domain of log x is (0,+∞). Implies sin(x-3) >0 & Domain of sin(x-3) is (-∞, +∞).But, now domain of sin(x-3) is (0, +∞). 0<(x-3)< ∞ . Now, add 3 to the inequality, it becomes 3<x<∞. Implies x>3 & x belongs to (3,∞).
Domain of f(x)= log[sin(x-3)] is (3,∞).
Finding domain of g(x)=√(16-x^2)
We know that domain of a √(x) is [0, +∞). Implies x>= 0. Here, in √(16-x^2) function, (16-x^2)must be >=0. (4-x)(4+x)>=0.
If (4-x)>=0 & (4+x)>=0 implies 4>=x(x<=4) &x>=-4 Now, x is [-4,4].
Domain of g(x) is [-4,4].
Now Resultant domain is common intervals of both f(x) &g(x). x belongs to (3,∞) & [-4,4]. It's common interval is (3,4].
Therefore Domain of Y is D(Y)= (3,4].