Inverse function of x^2?

[Willowz]

Find the inverse of the function ƒ(x) = x2. If not possible, explain why?


Relevant information. (http://en.wikipedia.org/wiki/Inverse_function#Example:_squaring_and_square_root _functions)

I drew up the function ƒ(x) = x2 on a graph and did the horizontal test. But, I do not understand why the horizontal test excludes attaining the inverse of that function.

[tiny-tim]

Hi Willowz! :smile:
… I do not understand why the horizontal test excludes attaining the inverse of that function.

ok … what's the inverse at f = 1 ? :wink:

(ie f-1(1))

[epenguin]

Does anyone else feel uncomfortable with tying up the student, who is 2nd year high school, in purely definitional problems apart from him getting stuck on it?

I do not know what you are required to answer, but in my book the inverse, f-1 of a function f, is the function that returns you to the value you started from, in other words f-1(f(x)) = x.

The horizontal test - I just looked up what that is, never needed a name for it before - tells you whether the inverse is unique. That is, does a given value of f(x) trace back only one or more than one value of x?

If more than one you can't talk of 'the' inverse though in my book you can talk of an inverse.

The inverse of squaring a number is finding its square root. symbolised √. But for heaven's sake that's only a word. √x just means the number whose square is x. So when you've said that in answer to a question like this you haven't really said much have you? Sure you're going to find √ all over the place in your math. But √(something) is only there all over the place because we know how to calculate it, given any number x. And how do you calculate it? Well you can probably do it for x = 36. Or x = 1, x = 4, x = 9 etc. You do it basically by knowing the answer. But if x is not such a special number, not an obvious 'perfect square', how do you do it? That is not so obvious is it? That is a little piece of real mathematics, as opposed to tiresome knowing definitions.

[epenguin]

I drew up the function ƒ(x) = x2 on a graph and did the horizontal test.

What values of x did you draw it between?

[Willowz]

What values of x did you draw it between?Do you mean for which y-values? For every y>0 it does not pass the test. I think your answer was sufficient. Thank you.

[osnarf]

If you were to graph the "inverse function" of f(x) = x^2 (flip it through the diagonal y = x), you'll see that the "inverse function" f-1(x) = +- sqrt(x) (see the bottom of the post if you dont see why it's +-). This represents a sideways parabola with center at (0,0), opening up to the right. So it's not defined for x < 0, and is equal to 0 for x = 0, which is all well and good.

However, for each x > 0, f-1(x) has two values, one positive and negative. In order to be a function, there can be only one number in the range associated with each number in the domain (in other words, in order to be a function it has to pass the vertical line test, which a sideways parabola clearly doesn't. In fact, anything which fails the horizontal line test will have an inverse that fails the vertical line test).

Because of this, we can't really even speak of the "inverse function", because what would be the inverse isn't a function.


---
f(x) = x^2
f-1(f(x)) = f-1(x^2)
x = f-1(x^2)

let x^2 = u
then x = +-sqrt(u)
for example, if u = 4, we need a number that, multiplied by itself, equals 4. 2 works, obviously, but since (-2)*(-2) = +4, -2 also works.

substitute x's for u's

+-sqrt(u) = f(u)
f(u) = +-sqrt(u)