x = -1y = 33x2 - y2 = 3 (-1)2 - (3)2 = 3 (1) - 9 = -6
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.
y = x This is a line and a function. Function values are y values.
f(f(x)) = f(x). Only if f is 1-1 then we have a solution f(x)=x.
x2 + y2 = 144 = radius squared radius = sqrt (144) = 12 diameter = 2 x radius = 24
Yes, as x-y2=0
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
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).
y2 + x = 7 : this can be written as y2 = 7 - x or x = 7 - y2. There is not a specific solution to this identity. For each different value assigned to x then a different value is generated for y. Or, it can be viewed that for each different value assigned to y then a different value is generated for x. This expression is therefore a function.
Yes
80
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
x2 - y2 = (x + y)(x - y) is an identity, not a function.
x = -1y = 33x2 - y2 = 3 (-1)2 - (3)2 = 3 (1) - 9 = -6
(x + y)(x - y) = 16
No. A simple example of this is y = x2; the inverse is x = y2, which is not a function.
2x2-y2