why doesn't f^-1(x)=square root of x, if f(x)=x^2?

Question

That is, why isn't the inverse of f(x)=x^2, equal to the square root of x?

Answer 1

If we are to take f as having a domain of the set of Real Numbers, then it is not one-to-one. Therefore, it is *not* invertible.

However, if we restricted the domain of f to only the non-negative numbers, and codomain to non-negative numbers, then f would be invertible, and sqrt would be its inverse.

It's a bit of an annoying technicality, but it really matters what the domain of the function is. You could define a function by using an expression that can be evaluated at all real numbers, but if you declare that the domain of a function f(x)=x^2 as being positive integers, then f(0) doesn't even exist; the function only takes positive integer inputs.

Answer 2

It does.

What you want to do is start with your equation f(x) = x^2. We can rewrite this equation in terms of x's and y's like this:

y = x^2

To find our inverse, we want to swap our x's and y's. So instead of:

y = x^2,

we have

x = y^2.

So, solving for y (again), we get that:

x = y^2
+-sqrt(x) = sqrt(y^2)
y = +-sqrt(x)

So it IS the square root of x. But technically, it has a plus or minus.

The only issue with that is whether or not the inverse is actually a FUNCTION. See, the way we determine whether the inverse of a function is a function or not is by using the HORIZONTAL LINE TEST (not the vertical. That's different). If we draw horizontal lines through the graph of f(x) = x^2, we see that any horizontal line is going to hit the graph twice which means the inverse we get (sqrt(x)) isn't actually a function.

So the inverse WOULD BE sqrt(x) but that inverse isn't actually a function.

That's the only issue I can see. Does that make sense?

Answer 3

The inverse of f(x) = 1/x^2
Therefore when x=2, the inverse of f(2) = 1/4
But, you are saying that the inverse of f(x) = sq rt (2) = 1.414
so, the inverse of f(x) is not the sq rt of x.

Answer 4

lets set f(x) = y, therefore y = x^2, the inverse would just be
x = sqrt(y), which is not the square root of x

Answer 5

It is not one-to-one
inverse function is symetric according to the line y=x