Q: Is the inverse of the function y equals x2 still a function?

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Y = 1/X2 ==============Can it pass the line test? * * * * * That is not the inverse, but the reciprocal. Not the same thing! The inverse is y = sqrt(x). Onless the range is resticted, the mapping is one-to-many and so not a function.

No. A simple example of this is y = x2; the inverse is x = y2, which is not a function.

Y = X2 Inverse. Y = 1/X2 ======

Yes

Y=X^2 is a function for it forms a parabola on a graph.

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Y = 1/X2 ==============Can it pass the line test? * * * * * That is not the inverse, but the reciprocal. Not the same thing! The inverse is y = sqrt(x). Onless the range is resticted, the mapping is one-to-many and so not a function.

No. A simple example of this is y = x2; the inverse is x = y2, which is not a function.

Y = X2 Inverse. Y = 1/X2 ======

y = x2 where the domain is the set of real numbers does not have an inverse, because the square root function is a one-two-two mapping (except at 0). Any polynomial with more than one root, over the reals, has no inverse. y = 1/x has no inverse across 0. But it is possible to define the domain so that each of these functions has an inverse. For example y = x2 where x is non-negative has the square root function as its inverse.

Yes. Think of y as being a function of x. y = f(x) = x2 + 1

No, it is not.

Yes

Y=X^2 is a function for it forms a parabola on a graph.

1 over x2 - 4 is the multiplicative inverse of x2 minus four 1/x2 - 4

The additive inverse of x2yz is -x2yz

Implicit: x2 + 2y = 5 Explicit : y = (5 - x2)/2

f(x) = x2 This describes a parabolic curve, with it's vertex at the point (0, 0)