MHB Why is y = x^2 not one-to-one?

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The function y = x² is not one-to-one because multiple inputs yield the same output; for example, both x = -1 and x = 1 result in y = 1. This can be demonstrated mathematically, as solving for x gives x = ±√y, indicating two x-values for any positive y. While y = x² passes the vertical line test, confirming it is a function, it fails the horizontal line test, which confirms it is not one-to-one. The discussion highlights the importance of domain restrictions to achieve a one-to-one function. Overall, y = x² is a valid function but does not meet the criteria for being one-to-one.
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Why is y = x^2 not one-to-one?
 
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Consider y = 1...there are two inputs (x = -1 and x = 1) that map to y = 1. And so y = x² is not one-to-one. :D

If we try to solve for x, we get:

$$x=\pm\sqrt{y}$$

This tells us that for a particular y greater than zero, we have 2 x's that map to it...unless we restrict x (the domain) such that it is either non-negative or non-positive.
 
Ok. If we let x = 1 or -1 for y = x^2, both values lead to y = 1 after squaring. We can also say that y goes to 1 for both values of x. The conclusion is that the parabola y = x^2 is a function but not one-to-one.

Correct?
 
RTCNTC said:
The conclusion is that the parabola y = x^2 is a function but not one-to-one.

Correct?

It passes the vertical line test, and so is a function, but fails the horizontal line test, and so is not one-to-one. :D
 
I like the vertical and horizontal line tests.

Question:

When does an expression fail to be a function?
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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