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How is sqrt(x) not a funtion?

  1. Aug 18, 2011 #1
    It's a question that has always popped up in my head. How is f(x)= ±[itex]\sqrt{x}[/itex] 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. jcsd
  3. Aug 18, 2011 #2
    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 [itex]f(x) = x^2[/itex] corresponds each real number x with its square, which is unique (i.e. each real number only has one square). However, note by symmetry, [itex]f(-2) = f(2)[/itex], and in general, [itex]f(-x) = f(x), \quad \forall x \in \mathbb{R}[/itex].

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

    However we can restrict the image of [itex]g(x) = \sqrt{x}[/itex] to only positive numbers so that [itex]g[/itex] returns a unique (positive) number for each number in its domain.
     
  4. Aug 18, 2011 #3
    So it mainly has to do with unique solutions? Also, does it also have to do with applications, such as Chemistry, Physics and Biology?
     
  5. Aug 19, 2011 #4

    chiro

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    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.
     
  6. Aug 21, 2011 #5
    It's an multivalued functions.
     
  7. Aug 21, 2011 #6

    HallsofIvy

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    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 [itex]y\ne y'[/itex] cannot be in the relation". If, in addition, you define the square root function to to include the pairs (x, y) as long as [itex]y^2= x[/itex], 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: " [itex]y= \sqrt{x}[/itex] 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 [itex]x^2= a[/itex]?" you would answer "[itex]x= \pm \sqrt{a}[/itex]" would you not? And then my question would be "If [itex]\sqrt{a}[/itex] means both the positive and negative roots, why do you need the "[itex]\pm[/itex]"?
     
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