No.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Both. If you look at it like this: y=41-x you say x is independent and y is dependent of x (i.e. y is a function of x) For x=41-y you say y is independent and x is dependent of y (i.e. x is a function of y)
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
The range depends on the domain, which is not specified.
y < 1
Since the range of the cosine function is (-1,1), the function y = cos(x) assumes a minimum value of -1 for y.
y = x This is a line and a function. Function values are y values.
Yes, but x would be a function of y, not the other (usual) way round. The domain of the function would be y in (-infinity, +infinity) and the range x in [0, +infinity).
The domain of a function represents the set of x values and the range represents the set of y values. Since y=x, the domain is the same as the range. In this case, they both are the set of all real numbers.
x2+2x+1=y or y=x2 In this function the domain is x equals real values and the range is y equals all real values provided y is more than or equal to zero.
If you plug in y for the x function, and it equals the answer you got, it is right.
Y = x squared -4x plus 3 is an equation of a function. It is neither called a domain nor a range.
The answer, for y as a function of x, depends on the range of y. Over the real numbers, it is not a function because a function cannot be one-to-many. But it is always possible to define the domain and range in such a way that the mapping in not one-to-many.
Yes, y=-2+x is a linear function.
y=|x|/4 The range is [0 , ∞ )