# How is sqrt(x) not a funtion?

1. Aug 18, 2011

### Bogrune

It's a question that has always popped up in my head. How is f(x)= ±$\sqrt{x}$ not a function, if both the solutions are "symmetrical?"
Does this also mean that the inverse fucntion of even or periodic functions aren't real functions either? Futhermore, does this somehow apply to three-variable graphs as well?

2. Aug 18, 2011

### Dr. Seafood

Functions map objects (arguments) in a set (domain) to a unique object (values) in another set (image).

Let's try to understand it this way: the function $f(x) = x^2$ corresponds each real number x with its square, which is unique (i.e. each real number only has one square). However, note by symmetry, $f(-2) = f(2)$, and in general, $f(-x) = f(x), \quad \forall x \in \mathbb{R}$.

Now consider the inverse "function" $f^{-1}(x)$. It corresponds each output of $f$ with its input, i.e. given $f(x)$, we need to find x. But we know that the outputs of the inverse function aren't unique: if $f(x) = 4$, we have either x = 2, or x = -2. These inverse outputs are not unique, thus the inverse of $f$ is not actually a function, by definition.

However we can restrict the image of $g(x) = \sqrt{x}$ to only positive numbers so that $g$ returns a unique (positive) number for each number in its domain.

3. Aug 18, 2011

### Bogrune

So it mainly has to do with unique solutions? Also, does it also have to do with applications, such as Chemistry, Physics and Biology?

4. Aug 19, 2011

### chiro

For a function that produces one output, the easiest way is a vertical line test. In your example with your +,- if you draw a vertical line at any point for x > 0, you will get two intersections: one for the positive and one for the negative (x = 0 doesn't have this problem).

Functions only produce a unique value, which is why any vertical line test will only give one intersection.

If you want to know whether a function has an "inverse" you do a horizontal line test. For example y = x has an inverse for all x, but y = x^2 does not because y = x^2 implies x = +,- SQRT(y) which is the same kind of problem you are describing. So if you wanted something to be a function, or have an inverse it has to pass the "vertical" or "horizontal" line test.

5. Aug 21, 2011

### simplicity123

It's an multivalued functions.

6. Aug 21, 2011

### HallsofIvy

As usual, the answer depends upon exactly how you define things. For real functions, the definition of "function" is normally, "a relation such that the pairs (x, y) and (x, y') for $y\ne y'$ cannot be in the relation". If, in addition, you define the square root function to to include the pairs (x, y) as long as $y^2= x$, it would include, for example, both (4, 2) and (4, -2) making it NOT function.

However, I would have to say that I would interpret "sqrt(x)" as the standard definition: " $y= \sqrt{x}$ is the non-negative number whose square is equal to x". With that definition, we have the (x, y) pair (4, 2) but not (4, -2) and that is a function.

If we were to ask "what numbers satisfy $x^2= a$?" you would answer "$x= \pm \sqrt{a}$" would you not? And then my question would be "If $\sqrt{a}$ means both the positive and negative roots, why do you need the "$\pm$"?