Is sqrt(x) a function if it includes both positive and negative solutions?

[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 function of even or periodic functions aren't real functions either? Futhermore, does this somehow apply to three-variable graphs as well?

 

[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.

 

[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?

 

[chiro]

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

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.

 

[simplicity123]

It's an multivalued functions.

 

[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 "$\pm$"?