The answer will range between '2' & '-2'
Reason; The Sine function ranges between '1' & '-1' , so if it has a coefficient of '2', this will increase the range to '2' & '-2'.
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y = 2sin(x)?
If that's your function, well we know that sin(x) oscillates between y = 1 and y = -1, but in our case we have double that from 2sin(x), so our range is -2 to 2.
x is a member of the function's domain, y is a member of the function's range.
The range is the y, while the domain is the x.
Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.
Domain (input or 'x' values): -∞ < x < ∞.Range (output or 'y' values): -2 ≤ y ≤ 2.
The inverse of the function y = x is denoted as y = x. The inverse function essentially swaps the roles of x and y, so the inverse of y = x is x = y. In other words, the inverse function of y = x is the function x = y.