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- az_lenderLv 72 months agoFavorite Answer
I see an unbalanced parenthesis after the "c" but I'm going to ignore it and suppose that the argument of the arcsin is (x/a + b). So the quantity (x/a + b) must be between -1 and +1, inclusive.
When x/a + b = -1, you have x/a = -b - 1, and x = -ab - a.
When x/a + b = +1, you have x/a = 1 - b, and x = a - ab.
The third choice in the list seems to be correct.
- PopeLv 72 months ago
You mean this?
f(x) = sin⁻¹(x/a + b) + c
a, b, c ∈ (-∞, 0)
The domain of the sin⁻¹ function is [-1, 1], and the added constant is of no consequence.
-1 ≤ x/a + b ≤ 1
-1 - b ≤ x/a ≤ 1 - b
-a - ab ≥ x ≥ a - ab ... (multiplication by a, a negative)
a - ab ≤ x ≤ -a - ab
a(1 - b) ≤ x ≤ -a(1 + b)
Domain: [a(1 - b), -a(1 + b)]
That is the second choice on the list.