Quote:
Originally Posted by Never Win
How do you know whether or not an equation is a function by just looking at the equation. For example, how do you know y^2= 4-x^2 is not a function. I know what the definition of a function is, I just don't understand how you figure this out when looking at an equation. Thanks in advance.
Intuitively, the idea is that a function f:X->Y maps each element of X to exactly one element of Y. More fundamentally, a function f:X-->Y is a set of ordered pairs (x,y), with x∈X and y∈Y, where (x,y)∈f means that f(x) = y. To satisfy the condition that each of point of X is mapped to exactly one element of Y, we impose the condition that if (x,y)∈f and (x,z)∈f, then it must be the case that y=z. (Otherwise, we would have f(x) = y and f(x) = z, where x would be getting mapped to two different points under f.)
Similarly, the solution set to an equation such as y^2 = 4-x^2 will be a set of ordered pairs. For example, the solution set to the equation you gave is {(x,y)∈ℝ^2 | y^2 = 4-x^2}. So, to determine whether or not the solution set to an equation defines a function, you have to determine whether or not for every (x,y) pair satisfying the equation, there is no other number y_2 such that (x,y_2) satisfies the equation.
You may have been taught a heuristic method for figuring this out that was called something like "the vertical line test", where you would graph the solution set to an equation and consider drawing vertical lines through it. If you could draw a vertical line that intersected the graph at more than one point, then the graph could not define a function. Going back to the earlier explanation, this is because you would be mapping an "x" to more than one "y".
For example, consider your equation: y^2 = 4-x^2
Rewriting this, we have x^2+y^2 = 4. This is simply the equation of a circle of radius 2 centered at the origin. Using the vertical line test, we see that each line intersects the graph of the solution set in two spots. This is because the square root function is multi-valued, as PTB points out. In particular, we observe that x=0, y=2 satisfies the equations, but so does x=0, y=-2. Consequently, the solution set of this equation cannot define a function.