Jay520 said:
Good. Just think of x^2 as a function of x, which I'm sure you already do. If you do that then yes, two different inputs (2 and -2) will have the same output (4), but that does not imply that they are equivalent. They just happen to have the same output when inputted in the quadratic function. Just like the sin function. The input 0 will have the same output as pi, which has the same output as 2pi, which has the same output as 3pi, 4pi, 5pi, etc. All of these have the same output (0). But no one would ever say 0 = pi = 2pi. No, but they would say sin(0) = sin(pi), etc Similarly, people just because X^2 = (-X)^2, that doesn't mean x = -x. It just mean the ^2 function is symmetrical across the Y axis and thus the inputs x & -x will have the same outputs, without being the same thing on their own. But you're tying to go the other way. When you go from 4 to x=2 or x=-2, you're going from the output to the input (because you're going from the squared value to the square root, or from y to x). If you do this, then you're finding the inverse of the quadratic function. And when you take the inverse of a function, you have to limit the domain to where it is one-to-one before doing so. So you have to limit the quadratic function to [0, infinty) or (-infinity, 0]. That way, the function you use will give you one result. Just like the sin function. If you want to take the inverse of a value, then you need to limit the domain to make the function one-to-one. If you didn't then it would be possible for you to input 1 into the inverse function, and get back pi/2, 3pi/2, 5pi/2, etc...implying that these values are all equal. But that's because you didn't restrict the sin function to where it was one-to-one. If you did then you would get one value. That's why you never see the square root function graph including both y and -y values. The square root function is an inverse of the quadratic function. And to get the inverse of the quadratic function, you must restrict the domain to either all positive or all negative values, so it's one-to-one. THEN, you can draw the inverse graph, which would only include one square root for every square. |
I agree that in order to evaluate the function 1/(x^2) you need to limit it to a non-zero evaluation for x^2, however you are making a major error.
What was given was this: x^2 = 4. The function is then f(x) = (x^2) - 4 not f(x) = (x^-2) - 4 (inverse function) so no limits need be imposed.
Resolving f(x)=(x^2) - 4 is simply a quadratic at y intercept -4, which has x intercepts at 2 and -2 confirming them as solutions.
